Department of
Computing Science
Genome Rearrangement Problems
David Alan Christie
Submitted for the degree of Doctor of Philosophy
to
The University of Glasgow,
August, 1998.
c
°1998,
David Alan Christie.
Abstract
Various global rearrangements of permutations, such as reversals and transpositions,
have recently become of interest because of their applications in computational molecular biology. A reversal is an operation that reverses the order of a substring of a
permutation. A transposition is an operation that swaps two adjacent substrings of a
permutation. The problem of determining the smallest number of reversals required
to transform a given permutation into the identity permutation is called sorting by
reversals. Similar problems can be defined for transpositions and other global rearrangements.
Related to sorting by reversals is the problem of establishing the reversal diameter.
The reversal diameter of Sn (the symmetric group on n elements) is the maximum
number of reversals required to sort a permutation of length n. Of course, diameter
problems can be posed for other global rearrangements.
These various problems are of interest because the permutations can be used to
represent sequences of genes in chromosomes, and the global rearrangements then represent evolutionary events. As a result, we call these problems genome rearrangement
problems.
Genome rearrangement problems seem to be unlike previously studied algorithmic problems on sequences, so new methods have had to be developed to deal with
them. These methods predominantly employ graphs to model permutation structure.
However, even using these methods, often a genome rearrangement problem has no obvious polynomial-time algorithm, and in some cases can be shown to be NP-hard. For
example, the problem of sorting by reversals is NP-hard, whereas the computational
complexity of sorting by transpositions is open. For problems like these, it is natural to seek polynomial-time approximation algorithms that achieve an approximation
guarantee.
In this thesis, we study several genome rearrangement problems as interesting and
challenging algorithmic problems in their own right, including some problems for which
the global rearrangement has no immediate biological equivalent. For example, we
define a block-interchange to be a rearrangement that swaps any two substrings of the
permutation. We examine, in particular, how the graph theoretic models relate to the
genome rearrangement problems that we study.
The major new results contained in this thesis are as follows:
• We present a 3/2-approximation algorithm for sorting by reversals. This is the
best known approximation algorithm for the problem, and improves upon the
7/4 approximation bound of the previous best algorithm.
Abstract
iii
• We give a polynomial-time algorithm for a significant special case of sorting by
reversals, thereby disproving a conjecture of Kececioglu and Sankoff, who had
suggested that this special case was likely to be NP-hard.
• We analyse the structure of the so-called cycle graph of a permutation in the
context of sorting by transpositions, and thereby gain a deeper insight into this
problem. Among the consequences are: a tighter lower bound for the problem,
a simpler 3/2-approximation algorithm than had previously been described, and
algorithms that, in empirical tests, almost always find the exact transposition
distance of random permutations.
• We introduce a natural generalisation of sorting by transpositions called sorting
by block-interchanges, and present a polynomial-time algorithm for this problem.
• We initiate the study of analogous problems on strings over a fixed length alphabet. We establish upper and lower bounds and diameter results for the problems
over a binary alphabet. We also prove that the problems analogous to sorting by
reversals and sorting by block-interchanges are NP-hard.
The thesis has the following structure:
In Chapter 1 we introduce several genome rearrangement problems, review the
existing literature on the algorithmic aspects of these problems, and summarise the
fundamental results and notations.
We study the problem of sorting by reversals in Chapter 2. At the start of this
chapter, we introduce a new graph, called the reversal graph of a permutation, that
turns out to be useful for finding sequences of reversals that sort the permutation. We
then use these graphs to obtain the 3/2-approximation algorithm, and the polynomialtime algorithm for the special case.
Chapter 3 contains a number of contributions that help to improve our understanding of the algorithmic issues involved in the problem of sorting by transpositions. We
note that an optimal sequence of transpositions never has to break apart elements i
and i + 1 once they are together in a permutation. Then we analyse the structure
of cycles in the cycle graph leading to a better understanding of how these cycles affect the general problem. This leads to a simpler 3/2-approximation algorithm for the
problem. We also obtain an improved lower bound for sorting by transpositions and
improve the lower bound for transposition diameter. This latter result is obtained by
describing an optimal sequence of transpositions that sorts the reverse permutation.
In Chapter 3 we also describe various approximation algorithms and an exact branch
and bound algorithm for sorting by transpositions. In empirical tests, described in this
chapter, we observe that our approximation algorithms require much fewer than 3/2
times the optimal number of transpositions to sort a random permutation (even though
we do not prove any approximation bound for the approximation algorithms we test).
We also observe that the exact algorithm could almost always find an optimal sequence
Abstract
iv
of transpositions for random permutations. We note that for most permutations the
optimal length of sequence of transpositions is equal to or very close to the lower bound.
Chapter 4 is devoted to the problem of sorting by block-interchanges. We describe
a sequence of block-interchanges that sorts a given permutation. We then show that
properties of the cycle graph, in the context of this problem, mean that the sequence
we describe is optimal. We also establish the block-interchange diameter of S n .
In Chapter 5 we study problems on strings, over a fixed length alphabet, that
are analogous to sorting by reversals, sorting by prefix-reversals (reversals that act on
prefixes), sorting by transpositions and sorting by block-interchanges. We adapt the
concept of the breakpoint, from problems on permutations, for use with these problems.
Then, for all of the problems, we obtain upper and lower bounds for the distance
between two binary strings, and we establish a diameter result on binary strings. We
also show that the problems analogous to sorting by reversals and sorting by blockinterchanges are NP-hard, even when the alphabet is binary.
Contents
Abstract
ii
Contents
v
List of Figures
viii
List of Tables
x
Preface
xi
1 Introduction and Background
1.1 Introduction . . . . . . . . . . . . . . . .
1.2 Sorting by reversals . . . . . . . . . . . .
1.3 Sorting signed permutations by reversals
1.4 Sorting by prefix-reversals . . . . . . . .
1.5 Sorting by restricted reversals . . . . . .
1.6 Sorting by transpositions . . . . . . . .
1.7 Sorting by restricted transpositions . . .
1.8 Sorting by block-interchanges . . . . . .
1.9 Sorting by reversals and transpositions .
1.10 Generating the symmetric group . . . .
1.11 Sorting by translocations . . . . . . . .
1.12 Syntenic edit distance . . . . . . . . . .
1.13 Other related problems . . . . . . . . . .
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2 Sorting by Reversals
2.1 Introduction . . . . . . . . . . . . . . . . . . .
2.2 Definitions . . . . . . . . . . . . . . . . . . . .
2.3 Reversal graphs simplify sorting by reversals .
2.4 A 3/2-approximation algorithm for sorting by
2.4.1 The 3/2-bound . . . . . . . . . . . . .
2.4.2 The cycle decomposition . . . . . . . .
2.4.3 The sequence of reversals . . . . . . .
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reversals
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vi
Contents
2.5
2.6
2.4.4 The algorithm . . . . . . . . .
2.4.5 Conclusion . . . . . . . . . . .
An easy case of sorting by reversals . .
2.5.1 2-reversals, exposed and hidden
2.5.2 Kernels . . . . . . . . . . . . .
Conclusion and open problems . . . .
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34
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46
3 Sorting by Transpositions
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Fundamental definitions and results . . . . . . . . . . .
3.2.1 Definitions . . . . . . . . . . . . . . . . . . . . .
3.2.2 Transposition equivalent permutations . . . . . .
3.2.3 The transposition cycle graph . . . . . . . . . . .
3.2.4 Even length cycles . . . . . . . . . . . . . . . . .
3.3 Fully oriented cycles . . . . . . . . . . . . . . . . . . . .
3.3.1 Fully oriented triples . . . . . . . . . . . . . . . .
3.3.2 Canonical labellings . . . . . . . . . . . . . . . .
3.3.3 Fully oriented cycles . . . . . . . . . . . . . . . .
3.3.4 Knots . . . . . . . . . . . . . . . . . . . . . . . .
3.3.5 Strongly oriented cycles . . . . . . . . . . . . . .
3.3.6 Super oriented cycles . . . . . . . . . . . . . . . .
3.4 A new 3/2-approximation algorithm . . . . . . . . . . .
3.5 A tighter lower bound . . . . . . . . . . . . . . . . . . .
3.5.1 Hurdles . . . . . . . . . . . . . . . . . . . . . . .
3.5.2 How tight are the lower bounds? . . . . . . . . .
3.5.3 The transposition diameter . . . . . . . . . . . .
3.6 Solving instances of sorting by transpositions in practice
3.6.1 The algorithms . . . . . . . . . . . . . . . . . . .
3.6.2 Permutations . . . . . . . . . . . . . . . . . . . .
3.6.3 How the algorithms performed . . . . . . . . . .
3.6.4 Analysis of the performance of the algorithms . .
3.7 Conclusion and open problems . . . . . . . . . . . . . .
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47
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4 Sorting by Block-Interchanges
4.1 Introduction . . . . . . . . . .
4.2 Minimal block-interchanges .
4.3 The graph model . . . . . . .
4.4 Block-interchange distance . .
4.5 Block-interchange diameter .
4.6 Conclusion . . . . . . . . . .
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vii
Contents
5 Sorting Strings by Global Transformations
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Reversal distance between binary strings . . . . . . . . . . . . . . . .
5.3.1 A lower bound . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.2 An upper bound . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.3 Reversal diameter of Bn . . . . . . . . . . . . . . . . . . . . .
5.3.4 Sorting by reversals . . . . . . . . . . . . . . . . . . . . . . .
5.4 Prefix-reversal distance between binary strings . . . . . . . . . . . .
5.4.1 A lower bound . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4.2 An upper bound . . . . . . . . . . . . . . . . . . . . . . . . .
5.4.3 Prefix-reversal diameter of Bn . . . . . . . . . . . . . . . . . .
5.4.4 Sorting by prefix-reversals . . . . . . . . . . . . . . . . . . . .
5.5 Transposition distance between binary strings . . . . . . . . . . . . .
5.5.1 A lower bound . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5.2 An upper bound . . . . . . . . . . . . . . . . . . . . . . . . .
5.5.3 Transposition diameter of Bn . . . . . . . . . . . . . . . . . .
5.5.4 Sorting by transpositions . . . . . . . . . . . . . . . . . . . .
5.6 Block-Interchange distance between binary strings . . . . . . . . . .
5.6.1 A lower bound . . . . . . . . . . . . . . . . . . . . . . . . . .
5.6.2 An upper bound . . . . . . . . . . . . . . . . . . . . . . . . .
5.6.3 Block-interchange diameter of Bn . . . . . . . . . . . . . . . .
5.6.4 Sorting by block-interchanges . . . . . . . . . . . . . . . . . .
5.7 NP-completeness of reversal distance and block-interchange distance
5.8 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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142
Glossary
144
Bibliography
147
List of Figures
1.1
1.2
1.3
1.4
1.5
1.6
1.7
Some example reversals. . . . . . . . . . . . . . . . .
Some example transpositions. . . . . . . . . . . . . .
Some examples of reversals on signed permutations.
Some example prefix-reversals. . . . . . . . . . . . .
Some example block-interchanges. . . . . . . . . . .
Some example translocations. . . . . . . . . . . . . .
An example syntenic edit distance problem. . . . . .
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2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2.19
2.20
An example cycle decomposition of rG0 (π). . . . . . . .
An example reversal graph . . . . . . . . . . . . . . . . .
The effect of ρ(u) when u is red. . . . . . . . . . . . . .
The effect of ρ(u) when u is blue. . . . . . . . . . . . . .
The resulting cycle decomposition. . . . . . . . . . . . .
The resulting reversal graph. . . . . . . . . . . . . . . .
A type 1 vertex. . . . . . . . . . . . . . . . . . . . . . .
A type 2 vertex. . . . . . . . . . . . . . . . . . . . . . .
Some short ladders. . . . . . . . . . . . . . . . . . . . . .
rG(π) for π = [0 4 2 6 8 7 3 5 1 9 14 15 12 13 10 11 16].
rF (π) for π = [0 4 2 6 8 7 3 5 1 9 14 15 12 13 10 11 16].
rL(M ) where M = {a, b, c, d, e} in Figure 2.11. . . . . .
The cycle decomposition C. . . . . . . . . . . . . . . . .
Part of a long ladder. . . . . . . . . . . . . . . . . . . . .
A short ladder. . . . . . . . . . . . . . . . . . . . . . . .
The 3/2-approximation algorithm . . . . . . . . . . . . .
Finding an elimination sequence for rR(C). . . . . . . .
rF ∗ (π3 ) . . . . . . . . . . . . . . . . . . . . . . . . . . .
rF ∗ (π4 ) . . . . . . . . . . . . . . . . . . . . . . . . . . .
rF ∗ (π5 ) . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3.1
3.2
3.3
3.4
tG(π),
tG(π),
How a
tG(π),
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54
54
for π = [3 4 1 2]. . . . . . . . . . . . .
for π = [6 1 5 7 12 4 11 2 9 3 8 10]. . .
transposition changes the cycle graph.
for π = [4 3 2 1]. . . . . . . . . . . . .
viii
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ix
List of Figures
3.5
3.6
3.7
3.8
3.9
3.10
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
3.20
3.21
3.22
3.23
3.24
3.25
4.1
4.2
4.3
4.4
4.5
How to sort a knot of length 5. . . . . . . . . . . . . . . . .
A sequence of transpositions that sorts ω5 . . . . . . . . . . .
The oriented cycle obtained by an applying an transposition
leaves with an unoriented cycle. . . . . . . . . . . . . . . . .
How C and D are interleaved initially. . . . . . . . . . . . .
How the cycles are interleaved after the 0-transposition. . .
How the cycle graph changes. . . . . . . . . . . . . . . . . .
How tG(π) changes as the result of the transposition. . . .
The first transposition of a (0, 2, 2)-sequence. . . . . . . . .
Special case of (d). . . . . . . . . . . . . . . . . . . . . . . .
Special case of (e). . . . . . . . . . . . . . . . . . . . . . . .
The new 3/2-approximation algorithm . . . . . . . . . . . .
Bafna and Pevzner’s 3/2-approximation algorithm . . . . .
A cycle graph with three components. . . . . . . . . . . . .
An optimal length sorting of κ2 . . . . . . . . . . . . . . . .
A sequence of transpositions that sorts R11 . . . . . . . . . .
The algorithm: approx . . . . . . . . . . . . . . . . . . . . .
The algorithm: tlis . . . . . . . . . . . . . . . . . . . . . .
The algorithm: random . . . . . . . . . . . . . . . . . . . . .
The algorithm: branch . . . . . . . . . . . . . . . . . . . . .
Performance of best. . . . . . . . . . . . . . . . . . . . . . .
A sequence of 13 transpositions that sorts π. . . . . . . . .
. . . . . . . 59
. . . . . . . 60
that inter. . . . . . . 65
. . . . . . . 66
. . . . . . . 67
. . . . . . . 69
. . . . . . . 70
. . . . . . . 72
. . . . . . . 73
. . . . . . . 74
. . . . . . . 75
. . . . . . . 76
. . . . . . . 78
. . . . . . . 80
. . . . . . . 83
. . . . . . . 85
. . . . . . . 87
. . . . . . . 88
. . . . . . . 90
. . . . . . . 103
. . . . . . . 104
An example of sorting by block-interchanges . . . . . . . . . . . . . . . .
The black edges that change when a block-interchange is applied: (a)
when the block-interchange is not a transposition, and (b) when the
block-interchange is a transposition. . . . . . . . . . . . . . . . . . . . .
How a minimal block-interchange changes tG(π). . . . . . . . . . . . . .
An algorithm to calculate bd(π). . . . . . . . . . . . . . . . . . . . . . .
An algorithm to perform an optimal sequence of block-interchanges. . .
108
109
111
112
112
List of Tables
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
3.20
3.21
3.22
3.23
3.24
random permutations, lower bound . . . . . .
random permutations, approx . . . . . . . . .
random permutations, alt approx
. . . . . .
random permutations, tlis . . . . . . . . . .
random permutations, random . . . . . . . . .
random permutations, rpt random
. . . . . .
random permutations, branch . . . . . . . . .
random permutations, best . . . . . . . . . .
3-cycle permutations, lower bound . . . . . . .
3-cycle permutations, approx . . . . . . . . .
3-cycle permutations, alt approx . . . . . . .
3-cycle permutations, tlis . . . . . . . . . . .
3-cycle permutations, random . . . . . . . . .
3-cycle permutations, rpt random . . . . . . .
3-cycle permutations, branch . . . . . . . . .
3-cycle permutations, best . . . . . . . . . . .
transposition tight permutations, lower bound
transposition tight permutations, approx . . .
transposition tight permutations, alt approx
transposition tight permutations, tlis . . . .
transposition tight permutations, random . . .
transposition tight permutations, rpt random
transposition tight permutations, branch . . .
transposition tight permutations, best . . . .
4.1
The number of permutations of size n that achieve bD(n). . . . . . . . . 113
5.1
5.2
The 18 possible combinations of S and T . . . . . . . . . . . . . . . . . . 133
The complexity of the different problems, and the diameter of Bn . . . . 143
x
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94
94
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99
Preface
Acknowledgements
First of all I would like to thank Rob Irving for his supervision. Rob’s guidance has
been crucial in getting this thesis into its current state. In particular, I am grateful for
the many helpful comments he has given that have helped to improve the quality of
my writing. Thanks too to everyone else who has been part of my supervisory team.
I would like to thank my family for all their support, especially in this final year.
I would like to thank EPSRC (the Engineering and Physical Sciences Research
Council) for their financial support in the form of a Research Studentship from October
1994 to September 1997.
Lastly, I would like to un-thank Partick Thistle for being relegated twice during
the course of my postgraduate studies.
Publications and papers
• R. W. Irving and D. A. Christie. Sorting by Reversals: on a conjecture of Kececioglu and Sankoff. Technical Report TR-95-12, Department of Computing
Science of The University of Glasgow, May 1995.
The contents of this paper are presented in Section 2.5. However, the presentation
of this work has been changed, so as to make it more consistent with the rest of
Chapter 2.
Tran [Tra97] subsequently and independently obtained an identical result to that
presented in this paper.
• D. A. Christie. Sorting permutations by block-interchanges. Information Processing Letters, 60(4):165–169, November 1996.
Chapter 4 is based on this paper.
• D. A. Christie. A 3/2-approximation algorithm for sorting by reversals. In Proceedings of the Ninth Annual ACM-SIAM Symposium on Discrete Algorithms,
San Francisco CA., pages 244–252, January 1998.
xi
Preface
xii
The contents of this paper are presented in Sections 2.3 and 2.4, with corrections
to a few minor errors in the paper.
• D. A. Christie and R. W. Irving. Sorting Strings by Global Transformations.
Submitted for publication.
This paper is based on parts of Chapter 5.
Statement of collaboration
The following summarises the role of my supervisor, Rob Irving, in work undertaken
jointly with him.
Chapter 2
The reversal graphs of Section 2.3 were developed from graphs in [IC95]. Rob is due
most credit for the graphs in [IC95], and suggested the method used to develop those
graphs into the form shown in Section 2.3. Rob proposed using the matching and
ladder graphs to generate the cycle decomposition used in Section 2.4. Section 2.5 was
originally ([IC95]) written jointly with Rob.
Chapter 3
The canonical labellings of cycles (Section 3.3.2) were suggested by Rob. He also
proposed Theorem 3.3.3. An initial proof of Theorem 3.5.5 was found by Rob, but the
proof presented in this thesis is quite different.
Chapter 5
The NP-completeness proofs in this chapter were developed with Rob.
Overlaps with work by other authors
Theorem 3.5.6 was proved independently by Meidanis, Walter and Dias, and this theorem appears as the main result in [MWD97b].
Declaration
This dissertation is submitted in accordance with the regulations for the degree of
Doctor of Philosophy in the University of Glasgow. No part of it has been previously
submitted by the author for a degree at any university and all results contained within
are claimed as original, except where indicated above.
Chapter 1
Introduction and Background
1.1
Introduction
It is common to compare strings by calculating the minimum number of operations
required to transform one into the other, where the set of allowable operations varies
according to the context. When the set of allowable operations consists of insertions,
deletions and substitutions, efficient algorithms to calculate such ‘edit distances’ are
well known , e.g., [WF74], and are described in most algorithms textbooks.
We say that insertions, deletions and substitutions are local operations. However,
it is possible to envisage global operations that can change the entire string in one
step. Global operations like this include the reversal and the transposition. A reversal
inverts the order of a substring of any length, and a transposition swaps two adjacent
substrings of any length. Examples of these global operations are shown in Figures
1.1 and 1.2. Note that in these figures, and in others too, the operations act on the
underlined substrings.
In this thesis we consider several edit distance problems based on global operations.
Motivation for studying these problems comes firstly from computational molecular
biology. At the chromosome level, genetic sequences mutate much more commonly
by global operations like reversals and transpositions, than by local operations. So
evolutionary relationships between genomes can be inferred by calculating such global
edit distances.
We call any problem to do with calculating edit distances based on global operations
76315824
76312854
76315824
13675824
Figure 1.1: Some example reversals.
1
Chapter 1. Introduction and Background
2
76315824
76312458
76315824
76158234
Figure 1.2: Some example transpositions.
a genome rearrangement problem. This definition is very broad, and includes problems
that are not directly motivated by biological applications, that involve more than two
strings, and that are based on sets instead of strings. Typically, however, we consider
genome rearrangement problems that involve calculating the edit distance between two
strings relative to one kind of global operation.
For each problem, we seek an efficient algorithm or, alternatively, a proof that
the problem is NP-hard. If we cannot obtain an efficient algorithm then we seek
good heuristics for solving instances of the problem, and efficient algorithms that find
provably good solutions. For minimisation problems, like all the problems we consider,
an algorithm is called an r-approximation algorithm if the solution produced by the
algorithm is guaranteed to be no greater than r times the optimal solution.
Genome rearrangement problems have been studied for only a relatively short while.
New methods have had to be developed for these problems, because the dynamic programming solutions typical of local edit distance algorithms are not obviously applicable
in the new setting.
One step that is almost always taken to simplify the problems is to assume that
the sequences being compared contain no repeated characters. The resulting simplified
problem is still biologically valid since a genome is unlikely to contain duplicate genes.
If there are no repeated characters then the strings are actually permutations, since
the global operations we consider cannot insert or delete any characters.
In this thesis we use π and φ to denote permutations. The length of a permutation
is denoted by n. (We also occasionally use |π| to denote the length of π.) We denote
the ith element of π by π(i). So, for example, if π = [4 1 3 2], then π(1) = 4, π(2) = 1,
and so on. For reasons that will become apparent later, we always assume that π is
extended so that π(0) = 0 and π(n + 1) = n + 1. However, usually when we write π, as
above, we do not include these extra elements. The identity permutation [1 2 3 . . . n]
is denoted by ın , or ı when the length of the permutation is obvious. The set (actually
the symmetric group) containing all permutations of length n is denoted by S n .
Global edit distance problems on permutations can be viewed as sorting problems.
For example, if π = [4 1 3 2], and φ = [2 4 1 3], we can relabel the elements of the
permutations so that φ = [ 1 2 3 4 ]. That is, we have relabelled 2 as 1 , 4 as 2 ,
1 as 3 , and 3 as 4 . With this relabelling π = [ 2 3 4 1 ], and finding the distance
between π and φ is equivalent to finding the shortest sequence of operations that sorts
π. Note that φ−1 · π = [2 3 4 1], and, in general, the distance between π and φ is equal
Chapter 1. Introduction and Background
3
to the distance between φ−1 · π and the identity permutation.
For a particular global operation X, sorting by Xs (e.g., sorting by reversals) is the
problem of finding a shortest sequence of Xs that sorts a given permutation. For such
a problem the distance of π, d(π), is the minimum number of operations required to
sort π. Note that by relabelling elements the distance between ı and π is equal to the
distance between π −1 and ı, so d(π) = d(π −1 ). The diameter of the symmetric group,
D(n), is the maximum value of d(π) taken over all n element permutations.
In this thesis we prefix d(π) with letters representing the global operation(s) being
considered. Similarly, in general, we prefix all functions by letters representing the
operation(s) being considered.
A simple concept that is often useful when studying genome rearrangement problems is that of the breakpoint. It is a position in the permutation that needs to be
changed by an operation, at some point in the transformation to the identity permutation. The exact definition of a breakpoint differs from problem to problem. A
lower bound for d(π) can be calculated from the number of breakpoints in π. Often
this lower bound can be improved by counting cycles in a graph, called the cycle graph,
that is particularly appropriate for capturing the ‘hidden’ difficulties that make genome
rearrangement problems non-trivial.
There are many different variations of genome rearrangement problems. One variation is derived from the fact that genes have an orientation in chromosomes. To model
this fact, elements of the permutations can be given signs, ‘+’ or ‘−’, that can change
when the elements take part in global operations. Other variations involve restricting
the size or positions of the rearrangements, e.g., to only allow reversals of length two.
Yet other variations involve different global operations to be described in detail later.
The genome rearrangement problems are interesting in an algorithmic sense, because some of the problems have been shown to be NP-hard, some are polynomial-time
solvable, and some have unresolved computational complexity. Sometimes changing a
problem’s definition just a little changes the problem from an NP-hard problem to a
problem in P, or vice-versa.
We can also view genome rearrangement problems as combinatorial puzzles like,
for example, the Rubik’s cube. In this way solving them is viewed as something of an
intellectual challenge, and motivation to solve the problems becomes intrinsic.
Recent textbooks by Setubal and Meidanis [SM97] and Gusfield [Gus97] both contain a chapter on genome rearrangement problems and, in particular, the problem of
sorting by reversals. In the sections below we describe specific genome rearrangement
problems and review the literature on these and related problems.
1.2
Sorting by reversals
A reversal ρ = ρ(i, j) (where 1 ≤ i < j ≤ n+1) transforms π into π ·ρ = [π(0) . . . π(i−
1) π(j − 1) . . . π(i) π(j) . . . π(n + 1)]. The reversal distance, rd(π), between π and
4
Chapter 1. Introduction and Background
ı is the length of a shortest sequence of reversals that transforms π into ı. Sorting by
reversals is the problem of finding a sequence of reversals of length rd(π) that sorts π.
When dealing with reversals, a (reversal) breakpoint is defined as a position i (1 ≤
i ≤ n + 1) such that |π(i) − π(i − 1)| 6= 1. (Remember that π is extended so that
π(0) = 0 and π(n + 1) = n + 1.) The number of reversal breakpoints in π is denoted by
rb(π). The identity permutation is the only permutation with no reversal breakpoints,
and a permutation can have no more than n + 1 reversal breakpoints. A strip is a
substring π[i..j] (i ≤ j) of π such that i and j + 1 are breakpoints, but no breakpoints
lie between these positions. A singleton is a strip of length one. A strip is increasing
if π(j) > π(i), otherwise it is decreasing.
A reversal can reduce the number of breakpoints in a permutation by at most two.
Therefore, a simple lower bound for reversal distance is
rd(π) ≥
rb(π)
.
2
Watterson, Ewens, Hall and Morgan [WEHM82] describe a simple approximation
algorithm for sorting by reversals1 that moves 1 to the front of the permutation with
the first reversal, then moves 2 to the correct place with the next reversal, and so on
until the permutation is sorted. This algorithm requires at most n − 1 reversals to
sort any given permutation. However, this algorithm does not achieve any constant
approximation bound.
Kececioglu and Sankoff [KS95]2 present a greedy approximation algorithm that
repeatedly applies a reversal that removes the most breakpoints from the permutation,
resolving ties among reversals that remove one breakpoint in favour a reversal that
leaves a decreasing strip. They are able to show that, on average, their algorithm
removes at least one breakpoint with every reversal, and is therefore a 2-approximation
algorithm. The worst-case time complexity of this algorithm is O(n2 ).
Bafna and Pevzner [BP96]3 introduce the important concept of the (reversal) cycle
graph 4 rG(π). It is an edge coloured undirected graph derived from π that contains
a vertex for each element in the permutation (including 0 and n + 1). So rG(π) has
vertex set {0, . . . , n + 1}. Vertices are joined by black edges if the elements they
represent form a breakpoint in π, i.e., rG(π) has black edge set {{π(i − 1), π(i)} :
1 ≤ i ≤ n + 1, i is a breakpoint}. Vertices i and i + 1 are joined by a grey edge if
these elements are not consecutive in π, i.e., rG(π) has grey edge set {{i, i + 1} : 0 ≤
i ≤ n, i and i + 1 are not consecutive in π}. Some example cycle graphs are shown in
Chapter 2.
The cycle graph is important because Bafna and Pevzner obtain an improved lower
bound for reversal distance by considering cycles in this graph. An alternating cycle is
1
In fact, they actually consider a similar problem on circular permutations in which mirror image
permutations are considered to be identical.
2
an earlier version of this paper appeared as [KS93]
3
an earlier version of this paper appeared as [BP93]
4
In fact, Bafna and Pevzner call these graphs breakpoint graphs.
Chapter 1. Introduction and Background
5
a cycle that has edges of alternating colours. The cycle graph can be completely decomposed into edge-disjoint alternating cycles, because each vertex has an equal number of
incident grey and black edges. The number of cycles in a maximum alternating-cycle
decomposition of rG(π) is denoted by rc(π). Bafna and Pevzner show that
rd(π) ≥ rb(π) − rc(π).
This is a better bound than the bound described earlier because each cycle accounts
for at least two breakpoints.
Bafna and Pevzner obtain a 7/4-approximation algorithm for sorting by reversals,
by considering signed permutations (Section 1.3) and properties of the cycle graph.
The worst-case time complexity of their algorithm is O(n2 ).
Christie [Chr98] presents a 3/2-approximation algorithm for sorting by reversals.
We describe this algorithm in Chapter 2. Frings [Fri98] describes experiments comparing this 3/2-approximation algorithm with Kececioglu and Sankoff’s 2-approximation
algorithm. He finds that, on average, the 3/2-approximation algorithm finds ‘significantly’ shorter sequences than the 2-approximation algorithm.
Hannenhalli and Pevzner [HP96] use their algorithm for sorting signed permutations
by reversals (Section 1.3) to prove a conjecture of Kececioglu and Sankoff [KS95] that
an optimal length sequence of reversals exists that sorts a permutation and does not
break apart strips of length three or more. They also explain precisely when strips of
length two must be split apart. In fact, for permutations that have fewer than log n
singletons they obtain a polynomial-time algorithm for sorting by reversals.
Kececioglu and Sankoff [KS95] describe some novel graphs that give improved lower
bounds for reversal distance that they use in an exact algorithm that can determine
the reversal distance of permutations of about 30 elements.
Caprara, Lancia and Ng [CLN95] describe an exact branch and bound algorithm
that sorts random permutations containing up to 100 elements in a few minutes.
Kececioglu and Sankoff [KS95] conjecture that deciding if a permutation can be
sorted with rb(π)/2 reversals is NP-complete and hence that sorting by reversals is NPhard. However, Irving and Christie [IC95], and Tran [Tra97] disprove this conjecture.
We explain this further in Section 2.5.
In fact, the general problem of sorting by reversals is NP-hard as proved by Caprara
[Cap97b] (see also [Cap97c]). Essentially, he obtains this result by proving that determining the value of rc(π) is NP-hard.
Caprara [Cap98] shows that the probability that the lower bound for rd(π) based
on rG(π) is not exact is low (asymptotically O(1/n5 )) for a random permutation π. He
uses this result to explain the good experimental performance of algorithms like that
of Caprara, Lancia and Ng.
The reversal diameter, rD(n), of the symmetric group Sn is the maximum value of
rd(π) taken over all permutations of length n. Bafna and Pevzner [BP96] prove that
rD(n) = n − 1 and that the only permutations needing this many reversals are the
Chapter 1. Introduction and Background
6
−7 −6 +3 +1 +5 +8 −9 −2 −4
−7 −6 +3 +1 +2 +9 −8 −5 −4
−7 +6 +3 −1 +5 +8 −9 −2 −4
+1 −3 −6 +7 +5 +8 −9 −2 −4
Figure 1.3: Some examples of reversals on signed permutations.
Gollan permutation, γn , and its inverse, where the Gollan permutation is defined as
γn+1 =
1.3


 [1],
if n = 0,
[γn (1) γn (2) . . . γn (n − 1) n + 1 γn (n)],
if n is odd,


[γn (1) γn (2) . . . γn (n − 2) n + 1 γn (n − 1) γn (n)], otherwise.
Sorting signed permutations by reversals
In fact, there is a more biologically relevant version of sorting by reversals in which every
element of the permutation is given a sign, ‘+’ or ‘−’, that indicates the orientation
of the gene in the chromosome. Such permutations are called signed permutations.
(Permutations with no signs are called unsigned permutations.) In this version of the
problem, a reversal flips the sign of all the elements it acts upon. Some examples of
reversals on signed permutations are shown in Figure 1.3.
The identity signed permutation is the permutation +ı = [+1 +2 . . . +n]. The
signed reversal distance srd(π) between π and +ı is the length of a shortest sequence
of reversals that transforms π into +ı. Sorting signed permutations by reversals is the
problem of determining a sequence of reversals of length srd(π) that sorts π.
For signed permutations, a signed reversal breakpoint is a position i (1 ≤ i ≤ n + 1)
such that π(i) − π(i − 1) 6= 1.
It would appear that sorting signed permutations by reversals is more complicated
than sorting by reversals, since any algorithm to solve the signed problem has to keep
track of the orientation of the elements as well as sort the permutation. However,
despite the apparent extra complication, Hannenhalli and Pevzner [HP95b] show that
the problem of sorting signed permutations by reversals is solvable in polynomial-time.
They achieve this result by first of all transforming the signed permutation π into an
unsigned permutation π 0 of length 2n by replacing +i by [2i−1 2i] and −i by [2i 2i−1].
Significantly, rG(π 0 ) has a unique cycle decomposition.
By considering how reversals can affect this unique cycle decomposition Hannenhalli
and Pevzner are able to show that
srd(π) = rb(π 0 ) − rc(π 0 ) + rh(π 0 ) + rf (π),
Chapter 1. Introduction and Background
7
where rh(π 0 ) is the number of so-called hurdles in the cycle graph, and rf (π) is 1 if π
is a so-called fortress and 0 otherwise. An excellent explanation of this result is given
by Setubal and Meidanis [SM97].
Berman and Hannenhalli [BH96] improve the O(n4 ) algorithm contained in [HP95b]
by producing an O(n2 α(n)) algorithm to solve the problem, where α is the inverse Ackerman function (a function that is unbounded but grows remarkably slowly). Kaplan,
Shamir and Tarjan [KST97] present an O(n2 ) algorithm that is also simpler. If the
value of srd(π) is required only, without a sequence of reversals, then Berman and
Hannenhalli’s method can be used in a O(nα(n)) algorithm.
Before Hannenhalli and Pevzner described their polynomial-time algorithm, Bafna
and Pevzner [BP96] had described a 3/2-approximation algorithm for the problem, and
Kececioglu and Sankoff [KS94] had observed that the lower bound based on the cycle
graph of π 0 was very tight.5
We denote by Σn the set of n element signed permutations. The signed reversal
diameter of Σn , srD(n), is the the maximum value of srd(π) taken over all π in Σn . For
n > 3, srD(n) = n + 1. When n is even, the diameter is achieved by the permutation
+Rn = [+n +(n − 1) . . . +1]. When n is odd and n > 3, the diameter is achieved by
the permutation [+2 +1 +3 +n +(n − 1) . . . +4]. These permutations are described
by Knuth in [Knu98] (Exercise 5.1.4–43).
Caprara [Cap97a] considers a generalisation of sorting signed permutations by reversals called multiple sorting by reversals. In an instance of multiple sorting by reversals we are given q signed permutations, π1 , . . . , πq , and we must construct a signed
permutation φ such that Σqi=1 srd(φ, πi ) is minimised, where srd(φ, πi ) is the signed
reversal distance between φ and πi . Caprara shows that multiple sorting by reversals
is NP-hard, even when q = 3.
1.4
Sorting by prefix-reversals
A problem related to sorting by reversals is the problem of sorting by prefix-reversals, alternatively known as the pancake problem because it was posed [Dwe75] in the
following way:
The chef in our place is sloppy, and when he prepares a stack of pancakes
they come out all different sizes. Therefore, when I deliver them to a
customer, on the way to the table I rearrange them (so the smallest winds
up on top, and so on, down to the largest on the bottom) by grabbing
several from the top and flipping them over, repeating this (varying the
number I flip) as many times as necessary. If there are n pancakes, what is
the maximum number of flips (as a function of n) that I shall ever have to
use to rearrange them.
5
In fact, Kececioglu and Sankoff consider a version of the problem on circular permutations.
Chapter 1. Introduction and Background
8
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Figure 1.4: Some example prefix-reversals.
A prefix-reversal is a reversal of the form ρ(1, j), for 2 ≤ j ≤ n + 1. Some example
prefix-reversals are shown in Figure 1.4. The prefix-reversal distance prd(π) of a permutation π is the minimum number of prefix-reversals necessary to sort the permutation.
Sorting by prefix-reversals is the problem of finding a sequence of prefix-reversals of
length prd(π) that sorts π.
We define prefix-reversal breakpoints in the same way as we defined reversal breakpoints except that we ignore the first position in the permutation since this value must
change with every prefix-reversal. So the total number of prefix-reversal breakpoints
prb(π) in π can be defined as
prb(π) =
(
rb(π),
if π(1) = 1,
rb(π) − 1, otherwise.
A prefix-reversal can decrease the number of prefix-reversal breakpoints by at most
one, so
prd(π) ≥ prb(π).
A simple algorithm to sort a permutation is to bring n to the front of the permutation with the first reversal, then move it to the end with the next reversal. We can then
use two reversals to move n − 1 to the correct place, and so on until the permutation
has been sorted. This process requires at most 2n − 3 reversals to sort π.
Heydari and Sudborough [HS] show that sorting by prefix-reversals is an NP-hard
problem.
The prefix-reversal diameter prD(n) of the symmetric group Sn is the maximum
value of prd(π) taken over all n-element permutations. Gates and Papadimitriou
[GP79] show that
5n + 5
.
prD(n) ≤
3
Heydari and Sudborough [HS97] improve a lower bound in [GP79] and show that
15n
≤ prD(n).
14
A signed version of sorting by prefix-reversals, also called the burnt pancake problem, is introduced in [GP79]. The signed prefix-reversal distance of π, sprd(π), is the
minimum number of prefix-reversals that sort a signed permutation π. The signed
Chapter 1. Introduction and Background
9
prefix-reversal diameter, sprD(n), of Σn is the maximum value of sprd(π) over all n
element signed permutations. Cohen and Blum [CB95] tighten bounds in [GP79] and
show that
3n
≤ sprD(n) ≤ 2n − 2.
2
Cohen and Blum also conjecture that the signed permutation prefix-reversal diameter
is achieved by the permutation −ı = [−1 −2 . . . −n]. Heydari and Sudborough [HS97]
show that sprd(−ı) ≤ 3(n + 1)/2, and that if the conjecture is true then prD(n) ≤
sprD(n) ≤ 3(n + 1)/2.
1.5
Sorting by restricted reversals
Chen and Skiena [CS96]6 investigate the problem of sorting permutations using reversals of a fixed-length only, i.e., using reversals of the form ρ(i, i + k) for some fixed k.
Note that sometimes it is impossible to sort a permutation using only reversals of a
certain length. For example, an odd length reversal does not change the parity of the
position of any element it acts on, so odd length reversals cannot sort any permutation
of the form π = [2 1 . . . ].
Chen and Skiena give a complete description for all n and k of how many permutations of length n can be sorted using only reversals of length k. They present upper
and lower bounds for the reversal diameter of the group of permutations of length n
that can be sorted by reversals of length k. They also study the problem on circular
permutations.
A reversal of length two simply swaps adjacent elements of π. Bubble sort is an
algorithm that sorts a permutation using the minimum number of reversals of length
two. The number of swaps it uses is the so called inversion number of the permutation.
Jerrum [Jer85] describes an algorithm that optimally sorts circular permutations using
only reversals of length two.
A short reversal is a reversal of the form ρ(i, i + 2) or ρ(i, i + 3). Vergara [Ver97]
studies the problem of sorting by short reversals. He presents a polynomial-time 2approximation algorithm for calculating the short reversal distance of a given permutation. He also presents bounds for the short reversal diameter of S n .
1.6
Sorting by transpositions
A transposition τ = τ (i, j, k) (where 1 ≤ i < j < k ≤ n + 1) transforms π into
π ·τ = [π(0) . . . π(i−1) π(j) . . . π(k −1) π(i) . . . π(j −1) π(k) . . . π(n+1)]. We view
a transposition as an operation that swaps two adjacent substrings in a permutation.
(An alternative and equivalent view is to think of a transposition as an operation
that removes the substring π[i..j − 1] from the permutation, before re-inserting this
6
an earlier version of this paper appeared as [CS95]
10
Chapter 1. Introduction and Background
substring in front of the element in position k.) Transpositions are also sometimes
called block-moves. (Note that a cycle of length two in the disjoint cycle form of a
permutation is sometimes called a transposition, but we are not concerned with these
kinds of transpositions.)
The transposition distance td(π) of a permutation π is the length of a shortest
sequence of transpositions that transforms π into ı. Sorting by transpositions is the
problem of finding a sequence of transpositions of length td(π) that sorts π. The
transposition diameter, tD(n), is the maximum value of td(π) taken over all n element
permutations.
When dealing with transpositions a (transposition) breakpoint is defined as a position i (1 ≤ i ≤ n + 1) such that π(i) − π(i − 1) 6= 1. (Remember, π is extended so
that π(0) = 0 and π(n + 1) = n + 1.) A strip is a substring π[i..j] of π (i < j) such
that i and j + 1 are breakpoints and there are no breakpoints between these positions.
The number of breakpoints in a permutation π is represented by tb(π). The identity
permutation is the only permutation that contains no breakpoints. A permutation can
contain at most n + 1 breakpoints.
A transposition can remove at most three breakpoints from a permutation. So we
can immediately obtain the following lower bound:
td(π) ≥
tb(π)
.
3
Bafna and Pevzner [BP98]7 introduce the important concept of the transposition
cycle graph for sorting by transpositions. The transposition cycle graph of π, tG(π), is
a directed edge-coloured graph with vertex set {0, . . . , n + 1}, grey edge set {(i, i + 1) :
0 ≤ i ≤ n}, and black edge set {(π(i), π(i − 1)) : 1 ≤ i ≤ n + 1}. Note that the edge
(u, v) is directed from u to v.
Since every incoming edge of a vertex can be uniquely paired with an outgoing edge
of the alternative colour, the graph can be completely decomposed into alternating
cycles, and furthermore this decomposition is unique. An alternating cycle is odd if it
contains an odd number of black edges. The number of odd alternating cycles in tG(π)
is denoted by tcodd (π).
The transposition cycle graph is important because Bafna and Pevzner show that
td(π) ≥
n + 1 − tcodd (π)
.
2
This is generally a much better lower bound for transposition distance than the bound
based on breakpoints. It is still an open question as to whether sorting by transpositions
can be solved in polynomial time, but Bafna and Pevzner use properties of cycles in the
transposition cycle graph to obtain a polynomial time 3/2-approximation algorithm for
the problem.
7
an earlier version of this paper appeared as [BP95b]
Chapter 1. Introduction and Background
11
The transposition diameter of the symmetric group, tD(n), is still unknown. Bafna
and Pevzner prove that
3n
n
≤ tD(n) ≤
.
2
4
They note that, for 3 ≤ n ≤ 10, tD(n) = bn/2c + 1, and that this value is achieved by
the reverse permutation Rn = [n n − 1 . . . 1]. They also state that td(Rn ) ≤ bn/2c + 1.
In Chapter 3 we prove that td(Rn ) = bn/2c + 1, and therefore tD(n) ≥ bn/2c + 1. This
result was obtained independently by Meidanis, Walter and Dias [MWD97b].
Guyer, Heath and Vergara [GHV95] describe heuristics for sorting by transpositions
that are based on the length of a longest increasing subsequence and the length of a
longest increasing substring of the permutation. In Chapter 3 we present evidence that
these heuristics are not very good.
Jordan [Jor95] makes an interesting connection between topology and sorting by
transpositions. He defines a surface based on tG(π), and shows that the lower bound
for td(π) based on tG(π) is the so-called genus of this surface.
1.7
Sorting by restricted transpositions
An insertion of the leading element is a transposition of the form τ (1, 2, i). Aigner
and West [AW87] investigate the problem of sorting a list by repeated insertion of the
leading element. They show that n − k insertions are required, where n is the length
of the list and k is the largest value such that the last k elements in the list form an
increasing sequence. This is proved by observing that the elements in positions n − k
and n − k + 1 must be in the wrong relative order at first. The only way their ordering
can change is if the element at position n − k reaches the front. This can only happen
after at least n − k − 1 insertions of the leading element. By extending the increasing
sequence at the end of the list with every insertion of the leading element, the list can
be sorted by n − k insertions.
An insertion of any element is a transposition of the form τ (i, i+1, j) or τ (i, j, j +1).
Knuth [Knu73] (Exercise 5.2.1–39), and Heath and Vergara [HV97] show that exactly
n − lis(π) insertions are required to sort π, where lis(π) is the length of a longest
increasing subsequence of π. The result is obtained because every element that is not
in the longest increasing subsequence must be moved, and every such element can be
moved into the correct position in the increasing subsequence by one insertion.
A bounded block-move is a transposition of the form τ (i, j, k), where k ≤ i + b
for some fixed parameter b. Heath and Vergara [HV97] investigate the problem of
sorting by bounded block-moves. A short block-move is a transposition of the form
τ (i, i + 1, i + 2), τ (i, i + 2, i + 3), or τ (i, i + 1, i + 3), i.e., a bounded block-move with
b = 3. Heath and Vergara [HV97] show that the short block-move diameter of S n is
¡ ¢
d n2 /2e. In [HV98] they obtain a polynomial-time 4/3-approximation algorithm for
calculating the short block-move distance of a given permutation. They also describe
Chapter 1. Introduction and Background
12
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Figure 1.5: Some example block-interchanges.
classes of permutations for which they can calculate the exact short block-move distance
in polynomial-time.
1.8
Sorting by block-interchanges
A block-interchange is a generalisation of a transposition. In a block-interchange two
non-intersecting substrings of any length are swapped in the permutation, whereas in
a transposition the substrings must be adjacent. Some example block-interchanges
are shown in Figure 1.5. The block-interchange distance of π, bd(π), is the length of
a shortest sequence of block-interchanges that transforms π into ı. Sorting by blockinterchanges is the problem of finding a sequence of block-interchanges of length bd(π)
that sorts π.
Christie [Chr96] introduces the problem of sorting by block-interchanges, describes
a polynomial time algorithm for sorting by block-interchanges, and determines the
block-interchange diameter of Sn . We present these results in Chapter 4.
1.9
Sorting by reversals and transpositions
All of the problems considered so far, allow only one kind of global rearrangement
operation. However, we can, of course, define problems in which more than one kind of
global rearrangement can be performed. For example, Sorting by reversals and transpositions is the problem of finding a shortest sequence of reversals and/or transpositions
that sorts a given permutation. Walter, Dias, and Meidanis [WDM98] investigate this
problem. It remains open as to whether the problem can be solved in polynomial time,
but they describe a 3-approximation algorithm for the problem.
They also investigate a signed version of sorting by reversals and transpositions.
For this problem they describe a 2-approximation algorithm, and show that the reversal
and transposition diameter of Σn is at least bn/2c + 2. Hannenhalli, Chappey, Koonin
and Pevzner [HCKP95] use exhaustive search to solve a particular instance of length
7 of sorting signed permutations by transpositions and reversals.
Gu, Peng and Sudborough [GPS96] investigate the the problem of sorting signed
permutations by reversals, transpositions and trans-reversals, where a trans-reversal is
a simultaneous transposition and reversal. More formally, the trans-reversal τ ρ(i, j, k)
Chapter 1. Introduction and Background
13
removes the substring π[i..j − 1] from π, reverses this substring, and inserts it in
front of the element in position k of π. They describe a polynomial time 2(1 + 1/k)approximation algorithm for their problem, where k ≥ 3 is any fixed integer 8 . Gu,
Peng, Iwata, and Chen [GPIC97] describe a simple greedy approximation algorithm
for this problem that in tests finds solutions that are very close to optimal.
Blanchette, Kunisawa and Sankoff [BKS96] conclude from experimental evidence
that a weighted version of sorting signed permutations by reversals and transpositions
in which transpositions are given roughly twice the weight of reversals is likely to give
more biologically meaningful results than the unweighted version.
1.10
Generating the symmetric group
Let g1 , . . . , gk be permutations of the set {1 . . . n}. The set containing all the
permutations that can be expressed as a product of these permutations is a group. We
say that g1 , . . . , gk are generators of this group. The size of the group that is generated
may be exponential in n and k. However, Furst, Hopcroft and Luks [FHL80] describe
a polynomial-time algorithm to test for membership of such a group.
Reversals, transpositions and other global rearrangements can be represented by
permutations, e.g., [3 2 1 4 . . . n] is a permutation that reverses the first three elements
of a permutation. Finding d(π) for a particular kind of rearrangement is equivalent
to finding a minimum length sequence of such permutations that produces π. Even
and Goldreich [EG81] show, by a transformation from the 3-exact-cover problem, that
the general problem of finding a minimal length sequence of generators that produce a
given permutation from a given set of generators is NP-hard. Jerrum [Jer85] has shown
that the problem is PSPACE-complete even when there are only two generators given.
However, in the problems that we study, the generator sets are fixed, i.e., they are
not part of the problem instances. This can make the problems more tractable. Some
computationally tractable fixed generator set problems are described in [Jer85].
1.11
Sorting by translocations
Organisms with more than one chromosome can evolve by genome rearrangements
called translocations. This leads to a genome rearrangement problem based on translocations.
A translocation as an operation that acts on two strings X and Y such that a
prefix X 0 of X is swapped with a prefix or a reversed suffix Y 0 of Y . Note that the
swapped strings can have different lengths, but must have length at least one and must
be shorter than the entire string (i.e. 1 ≤ |X 0 | ≤ |X| − 1 and 1 ≤ |Y 0 | ≤ |Y | − 1).
Some example translocations are shown in Figure 1.6. Sorting by translocations is the
8
In a forthcoming paper Gu, Peng and Sudborough [GPS99] describe a 2-approximation algorithm
for this problem.
Chapter 1. Introduction and Background
14
[5 6 4] [1 2 3 7 8]
[1 2 3 4] [5 6 7 8]
[8 7 3 4] [5 6 2 1]
[1 2 3 4] [5 6 7 8]
Figure 1.6: Some example translocations.
problem of finding a shortest sequence of of translocations, of length tld(A, B), that
transforms A into B, where A and B are sets of strings. In this problem it is assumed
that each element appears in exactly one string of A and in exactly one string of B.
We also consider strings X and Y to be identical, if X = Y or X = Y , where Y is the
reverse of Y .
Kececioglu and Ravi [KR95] describe a 2-approximation algorithm for sorting by
translocations, and a 2-approximation algorithm for sorting by translocations and reversals. Hannenhalli [Han96]9 describes a signed version of sorting by translocations.
He shows, using methods similar to those used to study sorting signed permutations
by reversals, that the signed version of the problem can be solved in polynomial-time.
The time complexity of the unsigned version remains open.
It is possible to extend the definition of a translocation to include operations that
act on empty prefixes and suffixes, as well as prefixes and suffixes that are the entire
string (i.e. to allow 0 ≤ |X 0 | ≤ |X| and 0 ≤ |Y 0 | ≤ |Y | in the definition of translocation
given above). With this definition of translocation, a fusion is a translocation that
joins two strings together, and a fission is an translocation that splits a single string
into two strings. Hannenhalli and Pevzner [HP95c] describe a polynomial algorithm
for a signed version of sorting by translocations, fusions, fissions and reversals.
1.12
Syntenic edit distance
Ferretti, Nadeau and Sankoff [FNS96] introduce a problem that is similar to sorting
by translocations except that in their problem translocations act on sets instead of
strings. Their motivation for this problem is that, in practice, the order of genes in a
chromosome is often unknown, whereas the chromosomal assignment of genes is known.
They define a translocation to be an operation that transforms sets X and Y into
(X − X 0 ) ∪ Y 0 and (Y − Y 0 ) ∪ X 0 , for some X 0 ⊆ X and Y 0 ⊆ Y . Note that if Y = {}
then the operation is a fission, and that if X = X 0 and Y 0 = {} then the operation is
a fusion.
The syntenic edit distance, sd(A, B), is the minimum number of these translocations
that can transform A into B, where A and B are sets of sets. We define |A| to be the
number of sets contained in A.
9
an earlier version of this paper appeared as [Han95a]
Chapter 1. Introduction and Background
15
{1, 2, 3 4} , {1, 2, 3 4} , {1, 2, 3 4}, {1, 2, 3 4}
{1, 2}, {3, 4}, {1, 2, 3 4} , {1, 2, 3 4}
{1, 2} , {3, 4}, {1, 2} , {3, 4}
{1}, {3, 4} , {2}, {3, 4}
{1}, {3}, {2}, {4}
Figure 1.7: An example syntenic edit distance problem.
There is a compact representation of the problem in which we relabel all the elements
that occur in the first set of A by 1, and all the elements that occur in the second
set of A by 2, and so on (removing any duplicates in the same set). Solving the
problem, in this form, requires a sequence of translocations that transforms B into
{{1}, {2}, . . . , {|A|}}. So in a sense the syntenic edit distance problem is a set
sorting problem rather than a permutation sorting problem. An example syntenic edit
distance problem and its solution is shown in Figure 1.7.
Ferreti, Nadeau and Sankoff [FNS96] present a simple heuristic for approximating
the syntenic edit distance. DasGupta, Jiang, Kannan, Li, and Sweedyk [DJK+ 97] show
that calculating syntenic edit distance is an NP-hard problem. They also describe a
simple 2-approximation algorithm for the problem.
1.13
Other related problems
Pevzner and Waterman [PW95] present a list of open problems in computational molecular biology that includes a section on genome rearrangement problems.
Lowrance and Wagner [LW75] study the problem of calculating the edit distance
between two strings where the set of allowable edit operations is extended to include
swapping two adjacent characters. They describe a polynomial algorithm for the problem when there are special restrictions on the weight of a swap operation. Wagner
[Wag83] shows that in general calculating the extended edit distance is NP-hard.
A mathematical puzzle that is somewhat related to sorting by reversals, is the
problem of reversing a train using a short spur line attached to the main track. This
problem is introduced by Dewdney [Dew87], and studied by Amato, Blum, Irani and
Rubinfeld [ABIR89] and Aggarwal and Leighton [AL90].
A shuffle operation on a deck of playing cards is similar to a rearrangement operation in a genome rearrangement problem. Aldous and Diaconis [AD86] and Diaconis,
McGrath and Pitman [DMP95] present statistical analyses of card shuffling techniques.
Similarly, twist operations on a Rubik’s cube are like rearrangements in genome
rearrangement problems. Eidswick [Eid86] presents some methods that are useful for
solving the Rubik’s cube and similar problems.
Chapter 2
Sorting by Reversals
2.1
Introduction
In this chapter we study the problem of sorting by reversals. The significant results of
this chapter are as follows:
• we define a graph called the reversal graph that is useful for finding sequences of
reversals that sort permutations (Section 2.3).
• we present a polynomial-time 3/2-approximation algorithm for sorting by reversals (Section 2.4). This is currently the best known approximation guarantee for
sorting by reversals.
• we show that it is possible to decide in polynomial-time whether a permutation
can be sorted using only reversals that remove two breakpoints (Section 2.5).
This disproves a conjecture of Kececioglu and Sankoff [KS95] that this special
case of sorting by reversals is NP-hard.
We begin the chapter with a section of standard definitions and results that are useful
for studying sorting by reversals.
2.2
Definitions
Let π be a permutation of length n. As is standard, we assume that π is extended so
that π(0) = 0 and π(n+1) = n+1, so that the first and last elements of the permutation
can be dealt with in the same way as all the other elements of the permutation. The
reversal ρ = ρ(i, j) (where 1 ≤ i < j ≤ n + 1) transforms π into π · ρ = [π(0) . . . π(i −
1) π(j − 1) . . . π(i) π(j) . . . π(n + 1)]. The reversal distance, rd(π), between π and
ı is the length of a shortest sequence of reversals that transforms π into ı. Sorting by
reversals is the problem of finding a sequence of reversals of length rd(π) that sorts π.
We say that π(i) is to the left of π(j) if i < j. Similarly π(i) is to the right of π(j)
if i > j. A reversal breakpoint is a position i in the permutation such that 1 ≤ i ≤ n + 1
16
Chapter 2. Sorting by Reversals
17
and |π(i) − π(i − 1)| 6= 1. The number of reversal breakpoints in a permutation π is
denoted by rb(π). A reversal adjacency is a position i in the permutation such that
1 ≤ i ≤ n + 1 and |π(i) − π(i − 1)| = 1. A strip is a substring π[i, j], (i ≤ j), of π
such that i and j + 1 are breakpoints, but no breakpoints lie between these positions.
Note that, for conciseness, in the rest of this chapter we may talk of breakpoints and
adjacencies where, of course, we mean reversal adjacencies and reversal breakpoints.
The identity permutation is the only permutation containing no breakpoints, and
a permutation of length n can have at most n + 1 breakpoints. For any reversal ρ,
rb(π) − rb(π · ρ) ∈ {−2, −1, 0, 1, 2}. A reversal ρ is a k-reversal if rb(π) − rb(π · ρ) = k.
The (reversal) cycle graph rG(π) is an edge coloured graph that was introduced
by Bafna and Pevzner [BP96]. The graph contains a vertex for each element in the
permutation (including 0 and n + 1). So rG(π) has vertex set {0, . . . , n + 1}.
Two vertices are joined by a black edge if the elements they represent form a breakpoint in π. Therefore rG(π) has black edge set {{π(i − 1), π(i)} : 1 ≤ i ≤ n +
1, i is a breakpoint}. Two vertices i and i + 1 are joined by a grey edge if these elements are not consecutive in π. Therefore rG(π) has grey edge set {{i, i + 1} : 0 ≤ i ≤
n, i and i + 1 are not consecutive in π}. In fact, the relationship between vertices of
rG(π) and the elements of π that they represent is so close that we often treat both
objects as if they were the same thing. For example, we often refer to elements of
rG(π) instead of vertices of rG(π).
An alternating cycle is a cycle that has edges of alternating colours. Henceforth all
cycles referred to in the cycle graph will be alternating cycles. The length of a cycle is
the number of black edges it contains. A k-cycle is a cycle of length k. A long cycle is
a cycle of length greater than two.
The cycle graph can be completely decomposed into edge-disjoint cycles, because
each vertex has an equal number of incident grey and black edges. However there are
likely to be many different such cycle decompositions of rG(π). The maximum number
of cycles in any cycle decomposition of rG(π) is denoted by rc(π). The cycle graph and
cycle decompositions are important because they give us the following lower bound for
rd(π).
Theorem 2.2.1 (Bafna and Pevzner) For any permutation π,
rd(π) ≥ rb(π) − rc(π).
2.3
Reversal graphs simplify sorting by reversals
In this section we show how to use a cycle decomposition C of rG(π) to find a sequence
of reversals that sorts π. To help us find the sequence of reversals we use another graph
called the reversal graph rR(C). Note that, of course, rR(C) does depend on π, but for
conciseness we do not show this in our notation.
Chapter 2. Sorting by Reversals
18
To construct the reversal graph we shall use the augmented cycle graph rG 0 (π).
This graph is constructed from rG(π) by adding black edges and grey edges between
adjacent elements of π, and directing all black edges in the graph from π(i) to π(i + 1).
In other words, rG0 (π) is an edge coloured graph with vertex set {0, . . . , n + 1}, grey
edge set {{i, i + 1} : 0 ≤ i ≤ n}, and black edge set {(π(i), π(i + 1)) : 0 ≤ i ≤ n}. For
a particular black edge (π(i), π(i + 1)), π(i) is said to be the tail, and π(i + 1) is said
to be the head.
A (alternating) cycle in rG0 (π) does not need to respect the directions of the black
edges. So the augmented cycle graph can be decomposed into cycles just like the cycle
graph. Note that an adjacency in π is a 1-cycle in rG0 (π). We can use this fact to
transform a cycle decomposition C of rG(π) into a cycle decomposition of rG 0 (π). Let
C + denote the cycle decomposition of rG0 (π) obtained from a cycle decomposition C of
rG(π) by adding 1-cycles to C.
Given a permutation π, and a particular cycle decomposition C of rG0 (π), we construct the reversal graph rR(C) as follows. We begin with the vertex set {u 0 , . . . , un }.
Each vertex in this set is associated with a grey edge of rG0 (π). The vertex associated
with grey edge {i, i + 1} is denoted by ui . We colour ui blue if {i, i + 1} is part of a
cycle, according to C, in which it connects the head of a black edge to the tail of a black
edge. Otherwise we colour ui red. The grey edge of rG0 (π) associated with a vertex u
of rR(C) is denoted by g(u).
Let u be a vertex in rR(C) such that g(u) is part of cycle C of the cycle decomposition C. We define lg (u) and rg (u) to be the positions in π of the leftmost and rightmost
elements, respectively, that are incident to g(u). So lg (ui ) = min(π −1 (i), π −1 (i + 1)),
and rg (ui ) = max(π −1 (i), π −1 (i + 1)). Similarly, we define lb (u) and rb (u) to be the positions of the leftmost and rightmost black edges, respectively, that are joined by g(u)
in C. (The position of a black edge is the position of the rightmost element it is incident
to.) Note that, unlike lg (u) and rg (u), the definitions of lb (u) and rb (u) depend on the
cycle decomposition C. However lg (u) ≤ lb (u) ≤ lg (u)+1, and rg (u) ≤ rb (u) ≤ rg (u)+1.
We connect vertices u and v of rR(C) with an edge if
(i)
(ii)
(iii)
(iv)
lb (u) < lb (v) < rb (u) < rb (v), or
lb (v) < lb (u) < rb (v) < rb (u), or
lg (u) < lg (v) < rg (u) < rg (v), or
lg (v) < lg (u) < rg (v) < rg (u).
Example A cycle decomposition of the augmented cycle graph of π = [7 5 6 3 2 4 1]
is shown in Figure 2.1. Given this cycle decomposition lg (u0 ) = 0, rg (u0 ) = 7, lg (u2 ) =
4, and rg (u2 ) = 5, whereas lb (u0 ) = 1, rb (u0 ) = 7, lb (u2 ) = 5, and rb (u2 ) = 5. The
reversal graph of this cycle decomposition is shown in Figure 2.2. Note that in this
figure vertex ui is given the label i. 2
Of course, reversal graphs are not unique for π, because cycle decompositions are
not unique.
19
Chapter 2. Sorting by Reversals
0
7
5
6
3
4
2
Figure 2.1: An example cycle decomposition of rG0 (π).
6
Blue
4
Red
5
Blue
3
Blue
1
Red
2
Blue
0
Blue
7
Blue
Figure 2.2: An example reversal graph
1
8
20
Chapter 2. Sorting by Reversals
(i)
(ii)
Figure 2.3: The effect of ρ(u) when u is red.
The reversal ρ(i, j) acts on the black edges (π(i − 1), π(i)) and (π(j − 1), π(j)) of
The augmented cycle graph of the resulting permutation is identical to rG 0 (π)
except that these two black edges are removed and replaced by (π(i − 1), π(j − 1)) and
(π(i), π(j)), and the direction is flipped of each black edge of the form (π(k), π(k + 1))
where i ≤ k ≤ j − 2.
Reversal graphs are given their name because each vertex u in the graph has a
reversal ρ(u) associated with it. We define ρ(u) to be the reversal that acts on the two
black edges of rG0 (π) that are joined by g(u) in C, where C is the cycle of C containing
g(u). If C is a 1-cycle, then ρ(u) is the identity reversal (i.e. the reversal that does
nothing).
Now let us consider the effect of applying ρ(u) when this reversal is not the identity
reversal. As was described above rG0 (π · ρ(u)) is identical to rG0 (π) except that two
black edges have been removed, two have been added, and the direction of some black
edges has been flipped. Clearly every cycle in C, except the cycle C containing the
removed black edges, exists in rG0 (π · ρ(u)). So we can find a cycle decomposition of
rG0 (π · ρ(u)) that contains all the cycles in C except C. The edges of rG0 (π · ρ(u)) that
are not part of these cycles form either one or two cycles depending on the colour of u.
If u is a red vertex then the edges of C now form two cycles in rG0 (π · ρ(u)). These
cycles are length 1 and length l − 1, where l is the length of C. This is illustrated
in Figure 2.3. In this figure g(u) is represented by a grey edge, and the other path
connecting the black edges joined by g(u) is represented by a dotted grey edge.
If u is a blue vertex then the edges of C form a cycle in rG0 (π · ρ(u)). This is
illustrated in Figure 2.4.
We denote by C · ρ(u) the cycle decomposition of rG0 (π · ρ(u)) that contains all
the cycles in C except C, replacing C with one or two cycles as described above. The
following lemma can be used to generate the reversal graph rR(C · ρ(u)) of π · ρ(u) from
rR(C).
rG0 (π).
Lemma 2.3.1 Let u be a vertex of rR(C) such that g(u) is part of cycle C. Then
rR(C · ρ(u)) can be derived from rR(C) by making the following changes to rR(C):
(i) For each vertex v (v 6= u), change the colour of v if and only if {u, v} is an edge in
21
Chapter 2. Sorting by Reversals
(i)
(ii)
Figure 2.4: The effect of ρ(u) when u is blue.
rR(C).
(ii) For each pair of vertices v and w in rR(C), such that {u, v} and {u, w} are edges
in rR(C), flip the adjacency of v and w.
(iii) If u is a red vertex, make it an isolated blue vertex.
Proof We first prove that these alterations to the graph are necessary.
(i) Suppose {u, v} is an edge of rR(C). Then ρ(u) changes the direction of one of the
black edges (in the augmented cycle graph) that is incident to g(v), but leaves the
direction of the other black edge alone. So the colour of v is different in rR(C · ρ(u))
than in rR(C).
(ii) Suppose that {u, v} and {u, w} are edges of rR(C). Then because the order of
elements of π between positions lb (u) and rb (u) − 1 (inclusive) are reversed by ρ(u) it
transpires that {v, w} is an edge of rR(C · ρ(u)) if and only if {v, w} is not an edge of
rR(C).
(iii) If u is red, then as shown in Figure 2.3, ρ(u) transforms rG0 (π) so that g(u) is
part of a 1-cycle. So u should be blue and isolated in rR(C · ρ(u)).
We now prove that no more alterations need to be made to the graph.
(i) If {u, v} is not an edge of rR(C) then the directions of both black edges incident to
g(v) remain the same or are both reversed by the action of ρ(u). Therefore the colour
of any vertex not adjacent to u should remain the same.
(ii) If {u, v} is not an edge of rR(C) then ρ(u) does not change which vertices are
adjacent to v. So the adjacencies of vertices not adjacent to u remain the same.
(iii) If u is blue then as shown in Figure 2.4, u is still blue in rR(C · ρ(u)). It is also
clear that {u, v} is an edge of rR(C · ρ(u)) if and only if {u, v} was an edge of rR(C).
2
From this lemma we see that, if u is a red vertex in rR(C), then we obtain rR(C·ρ(u))
from rR(C) by flipping the colour of every vertex adjacent to u, flipping the adjacency
22
Chapter 2. Sorting by Reversals
0
7
2
3
6
5
4
1
8
Figure 2.5: The resulting cycle decomposition.
6
Red
4
Blue
5
Blue
3
Red
1
Blue
2
Blue
0
Blue
7
Blue
Figure 2.6: The resulting reversal graph.
of every pair of vertices adjacent to u, and making u an isolated blue vertex.
Similarly, if u is a blue vertex then we obtain rR(C · ρ(u)) from rR(C) by flipping
the colour of every vertex adjacent to u, and flipping the adjacency of every pair of
vertices adjacent to u. Note that in this case u remains blue and is still adjacent to
the same vertices.
Example Applying the reversal represented by vertex 4 in Figure 2.2 results in
the cycle decomposition in Figure 2.5. The reversal graph of this cycle decomposition
is shown in Figure 2.6. 2
As a simple consequence of Lemma 2.3.1 we state the following corollary without
proof.
Corollary 2.3.1 A reversal represented by a vertex u in rR(C) affects only vertices
that are in the same connected component as u.
23
Chapter 2. Sorting by Reversals
The reversal graph for the identity permutation consists of n + 1 isolated blue
vertices. So to sort π, a sequence of reversals is required that transforms the reversal
graph into a graph with n + 1 isolated blue vertices. A sequence of reversals, each of
the form ρ(u) for some vertex u in rR(C), that transforms rR(C) into n + 1 isolated
blue vertices is called an elimination sequence for rR(C). Similarly, a sequence of
reversals, each of the form ρ(u) for some vertex u in rR(C), that transforms a connected
component of rR(C) containing k vertices into k isolated blue vertices is called an
elimination sequence for that component. We now show that there are two kinds of
component in rR(C) and we show how to eliminate each kind, using only reversals
represented in the graph. But first, we need some lemmas about the structure of
rR(C).
Cycles C and D of C interleave if rR(C) contains vertices u and v, arising from
cycles C and D respectively, such that {u, v} is an edge of the reversal graph. Grey
edges of rG0 (π) interleave if the vertices that represent them are adjacent in the reversal
graph.
Lemma 2.3.2 Isolated blue vertices in rR(C) correspond to 1-cycles in C.
Proof From the definition of cycle graphs it is easy to verify that all 1-cycles in C
have corresponding isolated blue vertices in rR(C).
Now, suppose, for a contradiction, that v is an isolated blue vertex that arises from
a cycle C in C of length greater than one. Then v represents a grey edge g that joins
two black edges of C. Name the elements of π incident to the grey edge, x and x 0 , and
the two remaining elements incident to these black edges, y and z. Since v is blue, π
can take only two forms as shown below.
π=
. . . x y . . . z x0 . . . (i)
. . . y x . . . x0 z . . . (ii)
Now consider an algorithm that visits each vertex of the augmented cycle graph
in the order 0, 1, . . ., n, n + 1. This algorithm follows each grey edge in the graph.
Now in order to visit the smallest value between x and x0 it must travel along a grey
edge that interleaves with g. But that would mean v was not isolated, so x must be
contiguous with x0 in π. But then the cycle containing the black edge (x, x0 ) must give
rise to vertices in rR(C) that are connected to v. This contradiction proves the lemma.
2
Let C be a cycle decomposition of rG0 (π). A cycle C of C is said to be unoriented
if every grey edge in the cycle connects the head of a black edge to the tail of a black
edge. Otherwise, the cycle is said to be oriented. (Caprara, perhaps more logically,
calls unoriented cycles directed cycles, and oriented cycles undirected, but we keep
to the more standard terminology.) Note that all vertices of rR(C) arising from an
unoriented cycle are blue.
24
Chapter 2. Sorting by Reversals
w
Red
u
v
Red
Red
Figure 2.7: A type 1 vertex.
Lemma 2.3.3 An unoriented 2-cycle of rG(π) must interleave with another cycle in
rG(π).
Proof Both of the grey edges in an unoriented 2-cycle connect a head to a tail, so both
of the vertices of rR(π) that represent the 2-cycle are blue. By the definition of rR(π)
these two vertices are not adjacent. Now by Lemma 2.3.2 the vertices are not isolated.
So the vertices must be adjacent to vertices arising from another cycle. Therefore, an
unoriented 2-cycle must interleave with another cycle. 2
A connected component of rR(C) is oriented if it contains a red vertex, or it consists
solely of an isolated blue vertex. Otherwise the component is unoriented. Let A be a
connected component of rR(C), and let u be a vertex in A. We define Au to be the
subgraph of rR(C · ρ(u)) that contains all the vertices of A.
Lemma 2.3.4 If a component A of rR(C) is oriented then it contains a red vertex u
such that every component of Au is oriented (or the component A consists of a single
isolated blue vertex).
Proof To set up a contradiction, suppose that A is not a single isolated blue vertex,
and that Au contains an unoriented component for every red vertex u in A. By the
definition of unoriented components, the unoriented component in Au cannot be an
isolated blue vertex. So in order that the unoriented component of Au contains two
blue vertices that are joined by an edge, it must be the case that, for every red vertex
u in A, there is a pair of distinct vertices v and w such that, either
1) v and w are both red, {u, v}, {u, w} are edges in A and {v, w} is not an edge in A
(Figure 2.7), or
2) v is red and w blue, {u, v}, {v, w} are edges in A and {u, w} is not an edge in A
(Figure 2.8).
Further, these vertices, v and w, are part of an unoriented component of A u .
Call u a type 1 vertex if only case 1) applies, and otherwise a type 2 vertex. Let
Vr be the set of red vertices in A, so that every vertex in Vr is either a type 1 or a
type 2 vertex. Define a mapping f : Vr → Vr by means of f (u) = v where v is defined
25
Chapter 2. Sorting by Reversals
u
v
w
Red
Red
Blue
Figure 2.8: A type 2 vertex.
by either case 1) or case 2) above, depending on whether v is a type 1 or a type 2
vertex. For a given red vertex u in A, define the sequence u0 = u, ui+1 = f (ui ) for
i ≥ 0. Then because the set Vr is finite, the sequence must cycle, i.e., there must be a
smallest integer k (k ≥ 2) such that uj = uj+k , for any j ≥ x, where x is some positive
integer.
Suppose that the cycle contains a type 2 vertex, us . Then there is a blue vertex
w such that {us+1 , w} is an edge and {us , w} a non-edge of A. But since us+2 is
in a unoriented component of Aus+1 , it follows that {us+2 , w} must be an edge of A.
Similarly, {us+3 , w}, {us+4 , w}, . . . must be edges of A, but this leads to a contradiction
since us = us+k , and {us , w} is a non-edge of A.
Therefore every vertex in the cycle is a type 1 vertex. Let us be a vertex in
the cycle. Since us is a type 1 vertex, there is a red vertex w such that {us , us+1 },
{us , w} are edges, and {us+1 , w} is a non-edge in A. Recall that us = us+k . Note
that it is impossible to have us+k−1 = w because then us would be connected to a red
vertex us+1 in Aus+k−1 . It follows that {us+k−1 , w} must be an edge of A. Similarly,
{us+k−2 , w}, {us+k−3 , w}, . . . must be edges of A, but this leads to a contradiction since
{us+1 , w} is a non-edge of A. 2
Lemma 2.3.5 Let B be an unoriented component of rR(C). Then Bv is an oriented
component for all vertices v in B.
Proof This is a simple consequence of Lemma 2.3.1. 2
It is possible to use Lemma 2.3.4 to find an elimination sequence for an oriented
component A of rR(C). We simply apply a reversal represented by a red vertex v, for
which all the components of Av are oriented, and repeat until A has been eliminated
completely. Note that this observation together with Lemma 2.3.5 mean that we can
always find an elimination sequence for rR(C).
A reversal ρ(u) is a k-move if the number of 1-cycles in C · ρ(u) is k greater than
in C. If u is a red vertex then ρ(u) is a 1-move or a 2-move. If u is a blue vertex then
ρ(u) is a 0-move.
The following simple proposition will help us prove an important lemma about
components of the reversal graph.
Proposition 2.3.1 Let v be a vertex in rR(C). Then ρ(v) is a 2-move if and only if
v arises from an oriented 2-cycle.
Chapter 2. Sorting by Reversals
26
Proof Suppose ρ(v) is a 2-move. Then C · ρ(v) contains two more 1-cycles than C by
Lemma 2.3.2. So v must be red because otherwise the lengths of cycles in C · ρ(v) are
no different from the lengths of cycles in C. Now because v is a red vertex, ρ(v) splits
a cycle into a 1-cycle and a cycle of length l − 1, where l is the length of the original
cycle. In order for the reversal to be a 2-move, l must be 2, because the lengths of the
other cycles are not altered by ρ(v). Therefore v arises from an oriented 2-cycle.
Clearly a reversal on an oriented 2-cycle is a 2-move. 2
Lemma 2.3.6 All the vertices that arise from the same cycle in C are part of the same
connected component of rR(C).
Proof Suppose, for a contradiction, that a cycle C in C, gives rise to vertices in rR(C)
that are in different components.
Now let us imagine an elimination sequence S for rR(C). Each reversal, ρ(u), in
this elimination sequence, where u is a red vertex arising from C, splits the cycle into a
1-cycle and a cycle of length l − 1. The last reversal of this kind, ρ(w), must therefore
be a 2-move, and it must increase the number of isolated blue vertices in the reversal
graph by two (Proposition 2.3.1). Let u be a vertex arising from C that is not part of
the same component as w.
By Corollary 2.3.1 two components of rR(C) never merge together as a result of
applying a reversal during the elimination sequence. So the elimination sequence S
defines elimination sequences for each of the components of rR(C), and u and w are
never part of the same component. Also by Corollary 2.3.1 the elimination sequence
on the component containing w can be applied before any of the other elimination
sequences. If we do this, then when we apply ρ(w) the number of isolated blue vertices
in the reversal graph must still increase by two (by Corollary 2.3.1). But by Proposition
2.3.1, C must be an oriented 2-cycle at that stage. However, by the definition of rR(C),
the two vertices representing an oriented 2-cycle are connected by an edge, whereas C is
represented by vertices u and w (at least) that do not share an edge. This contradiction
proves the lemma. 2
It is possible to use Lemma 2.3.4 and Lemma 2.3.6 to calculate how many reversals
are required to eliminate an oriented component of rR(C) as shown by the following
proposition.
Proposition 2.3.2 Let A be an oriented component of rR(C) that contains vertices
arising from k different cycles of rG(π). Then there is an elimination sequence for A
that contains k 2-moves with all the other reversals being 1-moves.
Proof The elimination sequence found by the method described above always applies
a reversal that is represented by a red vertex in the reversal graph. So each reversal is
a 1-move or a 2-move. By Proposition 2.3.1 k of the reversals are 2-moves. 2
Chapter 2. Sorting by Reversals
27
We now describe how to deal with unoriented components. To eliminate an unoriented component we first make it an oriented component as described in Lemma 2.3.5.
The resulting oriented component can then be solved as before. So the number of
reversals needed to eliminate an unoriented component of rR(C), using only reversals
represented in rR(C), is the number stated in the following proposition that can be
easily verified.
Proposition 2.3.3 Let A be an unoriented component of rR(C) that contains vertices
arising from k different cycles of rG(π). Then there is an elimination sequence for A
contains one 0-move, and k 2-moves, with all the other reversals being 1-moves.
We now combine the last two propositions to find an upper bound on the reversal
distance of the permutation.
Theorem 2.3.1 Let C be a cycle decomposition of rG(π). Then, using only reversals
represented in rR(C + ), π can be sorted by rb(π) − |C| + ru(C) reversals, where ru(C) is
the number of unoriented components in rR(C + ), and |C| is the number of cycles in C.
Proof By repeatedly applying Propositions 2.3.2 and 2.3.3 we will find a sequence
of reversals that sorts π and contains ru(C) 0-moves, |C| 2-moves, and rb(π) − 2|C|
1-moves. So the total length of this sequence is rb(π) − |C| + ru(C). 2
Note that it may be possible to sort π with fewer reversals by using a different cycle
decomposition, or by using reversals that act on black edges of two different cycles.
In fact the bound of Theorem 2.3.1 is very similar to the bound obtained by Hannenhalli and Pevzner [HP95b] for the related problem of sorting signed permutations
by reversals. Applied to unsigned permutations, they proved that, for a permutation
π and a cycle decomposition C,
rd(π) ≤ rb(π) − |C| + rh(C) + rf (C),
where rh(C) is the number of hurdles contained in C, and rf (π) = 0, or 1. Hurdles are
a subset of the unoriented components of rR(C), and in fact rh(C) + rf (C) ≤ ru(C).
So their bound is tighter than the bound obtained in Theorem 2.3.1.
However, the elimination sequences of Proposition 2.3.2 and Proposition 2.3.3 are
crucial for a counting argument used in a later proof that establishes the 3/2 bound
of an approximation algorithm. So we prefer not to use the bounds established in
[HP95b]. This preference also has the added advantage that in the next section, we do
not need to use properties of signed permutations in order to establish results that are
purely about unsigned permutations.
Chapter 2. Sorting by Reversals
2.4
28
A 3/2-approximation algorithm for sorting by reversals
In this section we describe a 3/2-approximation algorithm for sorting by reversals. The
rest of this section is split into five subsections. In the first of these subsections (Section
2.4.1) an upper bound is described that is a sufficient condition for an algorithm to
achieve a 3/2-approximation bound. A method of generating cycle decompositions is
presented in Section 2.4.2. These cycle decompositions have reversal graphs, as shown
in Section 2.4.3, that can be used to sort π with a sequence of reversals that achieves
the bound. All the ideas presented in the earlier sections are pulled together in Section
2.4.4, where the 3/2-approximation algorithm is presented. Some concluding remarks
appear in Section 2.4.5.
2.4.1
The 3/2-bound
In this section we describe a sufficient condition for an algorithm to achieve a 3/2approximation bound. In the later sections we set out to produce an algorithm that
satisfies this condition.
Theorem 2.4.1 If rc2 (π) is the minimum number of 2-cycles in any maximum cycle
decomposition of rG(π) then
2
1
rd(π) ≥ rb(π) − rc2 (π).
3
3
Proof Bafna and Pevzner [BP96] proved the fundamental lower bound
rd(π) ≥ rb(π) − rc(π).
(2.1)
Now let C be a maximum cycle decomposition of rG(π) with the minimum number
of 2-cycles. Suppose, in particular, that C contains rc3∗ (π) cycles of length greater than
two. Then, by (2.1),
rd(π) ≥ rb(π) − rc2 (π) − rc3∗ (π).
However, because each long cycle accounts for at least three of the black edges that
are not in shorter cycles of the decomposition, it must be that
1
rc3∗ (π) ≤ (rb(π) − 2rc2 (π)).
3
Combining these two inequalities proves the theorem. 2
In view of the above theorem it is apparent that an algorithm that sorts π using at
most rb(π) − rc2 (π)/2 reversals will achieve a 3/2-approximation bound for sorting by
reversals.
Chapter 2. Sorting by Reversals
2.4.2
29
The cycle decomposition
In Section 2.4.1 we obtained a bound for reversal distance based on the minimum
number of 2-cycles in a maximum cycle decomposition. Perhaps perversely, we now
seek a cycle decomposition that contains as many 2-cycles as possible.
We will use a new graph called the matching graph rF (π) to find our cycle decomposition of rG(π). The matching graph contains a vertex for each black edge in
rG(π). Two vertices, u and v, of rF (π) are connected by an edge if rG(π) has a 2-cycle
that contains both of the black edges represented by u and v. So each edge in rF (π)
corresponds to a 2-cycle in rG(π).
A maximum cardinality matching M of rF (π) can certainly be found in polynomialtime. (Algorithms for finding the maximum cardinality matching of a graph are described, for example, in Chapter 9 of [LP86]).
Unfortunately the cycles represented by M may not be edge-disjoint. For example,
if π = [0 9 3 4 6 5 8 7 1 10 2 11] then M contains edges representing the cycles
(0, 1, 10, 9) and (9, 10, 2, 3), but both these cycles involve the same grey edge (9, 10).
However, it is possible to find a cycle decomposition that contains at least d|M |/2e
2-cycles.
To find an appropriate set of edge-disjoint 2-cycles we use a graph derived from M ,
called the ladder graph rL(M ). Let rL(M ) be a graph with a vertex for every 2-cycle
represented in the matching M . Connect vertices of rL(M ) if the 2-cycles that they
represent have an edge in common.
Proposition 2.4.1 rL(M ) consists of isolated vertices and simple paths.
Proof Vertices of rF (π) represent black edges of rG(π), and edges of M represent
2-cycles of rG(π). So, if any 2-cycles represented in M share black edges then M could
not be a matching. So the 2-cycles represented by M can share only grey edges. A
2-cycle has two grey edges that could each be shared with a different cycle, so vertices
of rL(M ) have maximum degree two. If two 2-cycles share a grey edge, but not a black
edge then they must share two vertices of rG(π). So the two cycles must be as shown in
Figure 2.9. If we extend these figures with more 2-cycles, that share grey edges but not
black edges, then we end up with 2-cycles sandwiched between other 2-cycles. From
the way that the cycles are sandwiched together, it is clearly impossible for rL(M ) to
contain a cycle. 2
We call the 2-cycles represented by edges in M that are represented by isolated
vertices in rL(M ) independent 2-cycles. By contrast we call the other 2-cycles represented by edges in M ladder cycles, and the vertices of rL(M ) representing these cycles
ladder vertices. This is because we call a collection of 2-cycles of rG(π) represented by
vertices that form a path in rL(M ) a ladder. (We leave it to the reader to work out
why we call these structures ladders.) Two very short ladders are shown in Figure 2.9.
Note that either all the cycles in a ladder are oriented, or all the cycles are unoriented. (A cycle of rG(π) is oriented or unoriented if and only if the cycle is oriented or
Chapter 2. Sorting by Reversals
30
Figure 2.9: Some short ladders.
unoriented respectively in rG0 (π).) So we define a ladder to be oriented or unoriented
depending on whether the cycles it contains are oriented or not.
Let us suppose that rL(M ) contains z isolated vertices and y ladder vertices. Note
that |M | = y + z.
Theorem 2.4.2 Given a maximum cardinality matching M of rF (π) it is possible to
find a cycle decomposition C of rG(π) that contains at least dy/2e ladder 2-cycles and
z independent 2-cycles.
Proof Construct rL(M ) for the matching M . This graph consists of simple paths
and isolated vertices. Let C contain all the independent 2-cycles from rL(M ) and every
alternate cycle from each ladder. Clearly these 2-cycles are edge-disjoint. The rest of
C can be obtained by adding any cycle decomposition of the remaining edges of rG(π).
2
Example Suppose π = [0 4 2 6 8 7 3 5 1 9 14 15 12 13 10 11 16]. Then rG(π)
is the graph shown in Figure 2.10. The matching graph rF (π) of this permutation
is shown in Figure 2.11. A maximum cardinality matching of this graph is indicated
by the edges a, b, c, d, and e. The ladder graph induced by this matching is shown
in Figure 2.12. Vertices a, c, d, and e may be selected from this graph. The cycle
decomposition C found by selecting these 2-cycles is shown in Figure 2.13. 2
Let C be the cycle decomposition found in Theorem 2.4.2. A selected 2-cycle is a
2-cycle of rG(π) that is represented in M , and is part of C. The edges of rG(π) that
are in 2-cycles represented in M , but not selected cycles, are called spare edges.
31
Chapter 2. Sorting by Reversals
0
4
2
6
8
7
3
5
1
9
14
15
12
13
10
11
16
Figure 2.10: rG(π) for π = [0 4 2 6 8 7 3 5 1 9 14 15 12 13 10 11 16].
04
51
42
a
35
26
b
73
9 14
c
13 10
15 12
d
11 16
e
Figure 2.11: rF (π) for π = [0 4 2 6 8 7 3 5 1 9 14 15 12 13 10 11 16].
a
b
c
d
e
Figure 2.12: rL(M ) where M = {a, b, c, d, e} in Figure 2.11.
0
4
2
6
8
7
3
5
1
9
14
15
12
Figure 2.13: The cycle decomposition C.
13
10
11
16
Chapter 2. Sorting by Reversals
32
We now consider the reversal graph rR(C) of this cycle decomposition.
Lemma 2.4.1 Let C be an unoriented ladder 2-cycle of C. Then the vertices of rR(C + )
that represent C are part of a component that contains vertices representing a cycle that
is not a selected 2-cycle.
Proof Let C be an unoriented ladder 2-cycle in C.
Suppose that the ladder containing C consists of three or more 2-cycles. Then there
must be another selected ladder 2-cycle D such that two spare black edges x and y
connect the black edges of C and D in the fashion shown in Figure 2.14. (Note that
the roles of C and D can be interchanged but the following argument holds with only
minor modifications.)
Now consider the cycle E containing x. Note that E cannot be a 1-cycle because we
are dealing with rG(π) and not rG0 (π). Now E interleaves with C or D if it contains
a black edge that occurs: (i) before the leftmost edge of C, (ii) between the two black
edges of D, or (iii) after the rightmost black edge of C. So E intersects with C of D
unless it is a 2-cycle containing x and y. However such 2-cycles cannot exist because
rG(π) would then contain a cycle with only grey edges, and that is obviously impossible
from the definition of rG(π). Therefore E intersects with C or D.
By a similar argument we can show that F , the cycle containing y, intersects with
C or D. (Note that E and F may be the same cycle.)
Now suppose both of E and F interleave with D but not with C. Then by Lemma
2.3.3 a cycle G interleaves with C. Now G must have a black edge, a, between the two
black edges of C, and a black edge b either before the leftmost black edge of C or after
the rightmost black edge of C. But since G is not C, E or F , it must be the case that
a is between the two black edges of D. So G interleaves with D.
Therefore, by Lemma 2.3.6, vertices representing C are part of the same component
in rR(C + ) as vertices representing E and F which contain spare edges.
Suppose that the ladder containing C is a ladder of length two, as shown in Figure
2.15. (Note that the roles of C and the spare edge can be interchanged but the following
argument holds with only minor modifications.) Suppose that the cycle or cycles
containing the spare black edges of the ladder do not interleave with C. Then, in a
similar way as was explained above, a cycle that interleaves with C must interleave
with the cycle containing the spare grey edge of the ladder. So vertices representing C
are part of the same component in rR(C + ) as vertices representing a cycle containing
a spare edge.
The proof is completed by noting that cycles in C containing spare edges cannot be
selected 2-cycles. 2
2.4.3
The sequence of reversals
We now explain how to find a sequence of reversals that sorts π in no more than
rb(π) − rc2 (π)/2 reversals. We can assume by, Theorem 2.4.2, that we have a cycle
33
Chapter 2. Sorting by Reversals
C
D
x
y
Figure 2.14: Part of a long ladder.
C
Figure 2.15: A short ladder.
decomposition C of rG(π) that contains at least dy/2e 2-cycles that are part of ladders
and z independent 2-cycles.
Theorem 2.4.3 It is possible to sort π using no more than rb(π) − rc2 (π)/2 reversals.
Proof Suppose that k of the selected 2-cycles in C are part of oriented components
of rR(C + ). Then, by Proposition 2.3.2, there is a sequence of reversals that eliminates
all the oriented components and includes at least k 2-moves and no 0-moves.
Suppose that there are u unoriented components in rR(C + ) that contain vertices
representing cycles that are not selected 2-cycles, and that together these components
include vertices representing l of the selected 2-cycles in C. Then, by Proposition 2.3.3,
there is a sequence of reversals that eliminates these components and includes at least
l + u 2-moves and only u 0-moves.
Suppose that the remaining v unoriented components of rR(C + ), which, by Lemma
2.4.1, consist only of vertices representing independent selected 2-cycles, contain vertices representing m 2-cycles. Then, by Proposition 2.3.3, there is a sequence of reversals that eliminates these components and includes at least m 2-moves and only v
0-moves. Note that v ≤ bz/2c, since, by Lemma 2.3.3, every component of this kind
represents at least two cycles.
The sequence of reversals found by the steps described will sort π using at least
k + l + m + u 2-moves and only u + v 0-moves. So at most rb(π) − k − l − m + v
reversals are required to sort the permutation. But k + l + m − v ≥ dy/2e + z − v,
since k + l + m ≥ dy/2e + z. Further dy/2e + z − v ≥ dy/2e + dz/2e, since v ≤ bz/2c.
Now |M | ≥ rc2 (π) and y + z = |M |, so dy/2e + dz/2e ≥ rc2 (π)/2. This establishes the
theorem. 2
Chapter 2. Sorting by Reversals
34
algorithm approximation(π: Permutation) is
begin
construct the matching graph rF (π);
find a matching M for this graph;
construct C using this matching;
construct the reversal graph rR(C + );
find an elimination sequence for rR(C + );
end approximation;
Figure 2.16: The 3/2-approximation algorithm
Hence an algorithm that finds the elimination sequence described above is a 3/2approximation algorithm for sorting by reversals.
2.4.4
The algorithm
The approximation algorithm that has been described in the previous sections is outlined in Figure 2.16. The matching graph can be constructed in O(n) time. A max1
imum cardinality matching for any graph can be found in O(|V ||E| 2 ), where |V | is
the number of vertices and |E| is the number of edges in the graph. The matching
graph contains at most n edges and n vertices so, even without exploiting the special
3
structure of the matching graph, the matching M can be found in O(n 2 ) time. The
cycle decomposition can then be found in O(n) time. Constructing the reversal graph
can be performed in O(n2 ) time since there are at most O(n2 ) edges. The elimination
sequence for components of R(C) can be found as shown in Figure 2.17. Clearly finding the elimination sequence in this way can be achieved in O(n4 ) time. So the overall
time-complexity of the algorithm is O(n4 ).
Note that by using methods described by Kaplan, Shamir and Tarjan [KST97] for
the related problem of sorting signed permutations by reversals, it is possible to find
the elimination sequence more efficiently. In fact, using their method of finding the
elimination sequence requires only O(n2 ) time and so the overall time-complexity of
the algorithm can be reduced to O(n2 ).
2.4.5
Conclusion
The approximation algorithm we have described has an approximation ratio of 3/2.
In fact the algorithm does perform that badly for some permutations. For instance
if π = [3 4 1 2] then the algorithm requires 3 reversals, when 2 is the minimum
possible. We define a sequence of permutations, of increasing length: π1 = π, πi+1 =
πi ++ [6i − 1 6i 6i + 3 6i + 4 6i + 1 6i + 2], where i ≥ 1, and ++ represents concatenation.
Chapter 2. Sorting by Reversals
algorithm find an elimination sequence for rR(C) is
begin
for each unoriented component loop
apply a reversal represented by a blue vertex;
end loop;
while π 6= ı loop
r:= first red vertex;
f ound:= f alse;
while not f ound loop
generate rR(C · ρ(r));
count unoriented components in rR(C · ρ(r));
if there are no unoriented components then
π:= π · ρ(r);
C:= C · ρ(r);
f ound:= true;
else
r:= next red vertex;
end if ;
end loop;
end loop;
end find an elimination sequence for rR(C);
Figure 2.17: Finding an elimination sequence for rR(C).
35
Chapter 2. Sorting by Reversals
36
All these permutations require 3/2 times the minimum number of reversals to be sorted
by our algorithm. Hence the algorithm achieves this worst bound asymptotically too.
On the other hand there are permutations for which the lower bound of Theorem
2.4.1 is exactly 2/3 of the actual reversal distance. For instance if π = [5 6 3 4 1 2] then
rd(π) = 3, whereas rb(π) = 4 and rc2 (π) = 2, so that the lower bound has value 2 in this
case. In a similar way to the above, we can define a sequence of permutations which all
have this property. Of course the approximation algorithm must sort all permutations
like this optimally in order to achieve the approximation bound. Therefore unless we
improve the lower bound we will not be able to improve the approximation ratio.
A strange property of our 3/2-approximation algorithm is that when we are constructing the cycle decomposition C, any cycle decomposition of those edges that are not
part of the selected 2-cycles will suffice. In general, the larger the cycle decomposition
we take of these edges, the better the solution we will find.
Similarly, a 0-reversal is used in the elimination sequence of every unoriented component of rR(C). However, in general, by using reversals that act on black edges of
two different cycles it is not necessary to use so many 0-reversals.
Frings [Fri98] has implemented the 3/2-approximation algorithm. He has tested the
algorithm with permutations of length 10, 20 and 40, that were generated by applying
k reversals to the identity permutation. In comparison with Kececioglu and Sankoff’s
2-approximation algorithm he discovered that 3/2-approximation algorithm found ‘significantly’ shorter sequences of reversals on average. For example, on permutations
with n = 20 and k = 7 the 3/2-approximation algorithm found an optimal length
sequence of reversals nearly 80% of the time, whereas the 2-approximation algorithm
found an optimal length sequence less than 60% of the time. However, there were some
permutations for which 2-approximation found shorter sequences of reversals than the
3/2-approximation algorithm. For example the 2-approximation algorithm uses only
two reversals to sort π = [3 4 1 2], but the 3/2-approximation algorithm uses three
reversals.
2.5
An easy case of sorting by reversals
Since the identity permutation has no breakpoints, and a reversal can reduce the number of breakpoints by at most two, an immediate lower bound for the reversal distance
is
rb(π)
.
(2.2)
rd(π) ≥
2
Let us call a permutation π reversal-tight if (2.2) is satisfied with equality. Kececioglu and Sankoff [KS95] made the following conjecture.
Conjecture 2.5.1 The problem of determining whether a given permutation is reversal-tight is NP-complete.
Chapter 2. Sorting by Reversals
37
In this section we describe a polynomial-time algorithm to determine whether a
permutation is reversal-tight, and thereby disprove the Kececioglu/Sankoff conjecture.
In fact, the same result has been obtained independently by Tran [Tra97].
2.5.1
2-reversals, exposed and hidden
A permutation π has an exposed 2-reversal ρ = ρ(i, j) (1 ≤ i, j ≤ n + 1, i < j − 1) if
and only if there are breakpoints at positions i and j, |π(i − 1) − π(j − 1)| = 1, and
|π(i) − π(j)| = 1. It is clear that application the reversal ρ on π will reduce the number
of breakpoints by 2 if and only if ρ is an exposed 2-reversal. (Exposed 2-reversals
correspond to oriented 2-cycles in rG(π).)
A permutation π has a hidden 2-reversal ρ = ρ(i, j) (1 ≤ i, j ≤ n + 1, i < j − 1)
if and only if there are breakpoints at positions i and j, |π(i − 1) − π(j)| = 1 and
|π(i) − π(j − 1)| = 1. Application of a hidden 2-reversal does not reduce the number
of breakpoints by 2. However, a hidden 2-reversal may be transformed to an exposed
2-reversal by the application of one or more other reversals, depending (in a precise
way to be elaborated upon below) on the overlapping pattern of the reversals. Hidden
2-reversals correspond to unoriented 2-cycles in rG(π). In what follows, we use the
term 2-reversal to mean either an exposed or a hidden 2-reversal.
Let R(π) be the set of exposed and hidden 2-reversals in π. At any stage during
the process of sorting π by reversals, application of an exposed 2-reversal cannot create
any new exposed or hidden 2-reversals — it will destroy any 2-reversal that shares an
end-point with it, and it may flip some exposed 2-reversals so that they become hidden,
and vice versa. So the lower bound in (2.2) can be achieved only if R(π) contains an
appropriate subset R0 (π) of exactly rb(π)/2 2-reversals. Here, ‘appropriate’ means:
(a) R0 (π) contains exactly one exposed or hidden 2-reversal with an endpoint at breakpoint position i, for each breakpoint position i in π; and
(b) the 2-reversals in R0 (π) can be ordered, say ρ1 , ρ2 , . . . , ρr (r = rb(π)/2)
so that, for each j (1 ≤ j ≤ r), ρj is exposed after ρ1 , . . . , ρj−1 have been
applied (in that order).
(Note that, when a reversal is applied, the ‘parameters’, or end positions, of any overlapping reversal will be changed, but we continue to think of it as essentially the same
reversal.)
Each of conditions (a) and (b) above can be expressed in terms of a (different)
graph model. For condition (a), we use the matching graph rF (π) of Section 2.4.2,
that has a vertex for each breakpoint i in π, with vertices i and j (i < j) adjacent
if and only if ρ(i, j) is a 2-reversal in R(π). (Note that in Section 2.4.2, rF (π) was
defined in terms of cycles of length 2 in rG(π) instead of 2-reversals. However, both
definitions give rise to the same graph.) Then condition (a) is clearly satisfied if and
only if the graph rF (π) has a perfect matching. So this is a necessary condition for π
Chapter 2. Sorting by Reversals
38
to be reversal-tight (and can easily be checked in polynomial time). That it is not a
sufficient condition may be seen from the second of the following illustrative examples.
Example
π1 = [0 9 6 4 2 8 1 7 3 5 10].
π1 has 10 breakpoints, and has two exposed 2-reversals — ρ1 = ρ(3, 8) and ρ2 = ρ(5, 7),
and four hidden 2-reversals — ρ3 = ρ(1, 6), ρ4 = ρ(2, 10), ρ5 = ρ(3, 9), and ρ6 = ρ(4, 9).
ρ1 , ρ2 , ρ3 , ρ4 , and ρ6 form a perfect matching in the graph rF (π1 ). The reversals may
be applied in the order ρ2 , ρ3 , ρ4 , ρ1 , ρ6 , so the reversal distance of π1 is 5, and π1 is
reversal-tight. 2
Example
π2 = [0 5 3 4 1 2 6 8 7 9].
π2 has 6 breakpoints, and has one exposed reversal — ρ1 = ρ(7, 9), and two hidden 2reversals — ρ2 = ρ(1, 4) and ρ3 = ρ(2, 6), and these 2-reversals form a perfect matching
in the graph rF (π). However, application of ρ1 leaves no exposed 2-reversal, so π2 is
not reversal-tight. 2
From now on we assume that the graph rF (π) has a perfect matching, since if it
does not, we can be sure that the permutation π is not reversal-tight. To investigate
condition (b) we use reversal graphs.
Let M be a perfect matching for rF (π). This perfect matching defines a cycle
decomposition CM of rG(π), in which all the cycles are 2-cycles. Our disproof of the
Kececioglu/Sankoff conjecture is based on the following theorem.
Theorem 2.5.1 A permutation π is reversal-tight if and only if there is some perfect
+
) is oriented.
matching M in rF (π) such that every component of rR(CM
+
Proof If M is a perfect matching in rF (π) such that every component of rR(CM
)
is oriented, then the elimination sequence found by using Proposition 2.3.2 on each
component sorts π such that every reversal is a 2-move. Further, each reversal must
remove 2-breakpoints because CM contains only 2-cycles. So π is reversal tight.
If π is reversal tight, then there must be a sequence of 2-reversals that sorts π. Each
2-reversal corresponds to a 2-cycle in rG(π). So rF (π) has a perfect matching. Further
+
each component of rR(CM
) is oriented, because the sequence of 2-reversals that sorts
+
π corresponds to an elimination sequence of rR(CM
), and none of the reversals are
0-moves. So each reversal corresponds to applying the reversal represented by a red
+
) then we
vertex of the reversal graph. If there was an unoriented component in rR(C M
would not be able to sort π. 2
2.5.2
Kernels
Theorem 2.5.1 on its own does not give a polynomial-time algorithm to check whether
a permutation is reversal-tight. Certainly, for a given permutation π we can check
Chapter 2. Sorting by Reversals
39
in polynomial time whether the graph rF (π) has a perfect matching M , and for any
such perfect matching M we can check whether each component of rR(CM ) is oriented.
But there may be many such matchings M — exponentially many in the worst case
— and we can only conclude that π is not reversal tight if rR(CM ) has an unoriented
component for every such matching M . We now investigate how multiple matchings
can arise, and how such situations can be handled.
Let π be a permutation and consider the graph rF (π). Recall that we are assuming
that the graph rF (π) has a perfect matching. In forming a perfect matching in this
graph, any vertex of degree 1 must be matched with the unique vertex to which it
is adjacent. We may therefore iteratively pair off vertices until we reach a subgraph
of rF (π) in which every vertex has degree ≥ 2. Furthermore, if this subgraph has a
bridge, i.e., an edge whose removal disconnects the graph, then just by considering the
parity of the two resulting components we can decide whether that edge is forced into
or out of every perfect matching. Dealing with every bridge in this way leads us to
a subgraph of rF (π) — say rF ∗ (π) — in which every vertex has degree 2 or 3, and
which contains no bridges. Let us call each connected component of rF ∗ (π) a kernel
in rF (π). It turns out that kernels have a very special structure, as will be revealed in
the following sequence of lemmas. (Of course, since we are assuming that rF (π) has a
perfect matching, each kernel necessarily has an even number of vertices.)
Vertices of rF (π) are associated with black edges of rG(π) which are in-turn associated with breakpoints of π. We shall blur the distinction between vertices of rF (π)
and breakpoints of π. So we say i is a breakpoint of a kernel K, if K contains a vertex
that is associated with breakpoint i of π. We use notation vi to denote the vertex of
K that is associated with the breakpoint of π at position i.
We need some additional terminology. Suppose that permutation π has a breakpoint i, and this breakpoint is represented by a vertex, v say, in the graph rF (π). We
call π(i − 1) and π(i) the components of vertex v. If {v, w} is an edge in a kernel K
then this edge is an {α, α + 1}-connection if α is a component of v and α + 1 is a
component of w, or vice-versa.
Call an element α of π one-sided with respect to kernel K if there is a breakpoint
of K immediately to one side of α but not to the other. An element of π is two-sided
with respect to K if it has breakpoints of K on both sides.
Suppose {v, w} is an edge of a kernel K, and that v and w represent, respectively,
breakpoints in positions i and j of the permutation, with i < j. Then v is called a
left breakpoint and w a right breakpoint with respect to K. (Note that later we show
that a breakpoint cannot be both a left and a right breakpoint with respect to K.)
For convenience, in what follows, we use α0 and α∗ to represent α + 1 and α − 1, not
necessarily respectively, for any element α of permutation π. If α0 represents α + 1 then
α00 represents α + 2, and so on.
Lemma 2.5.1 For all permutations π, if a kernel K of the graph rF (π) contains an
{α, α + 1}-connection, then K contains at least two {α, α + 1}-connections.
Chapter 2. Sorting by Reversals
40
Proof Suppose that the end-vertices of the given {α, α + 1}-connection are v and w.
Without loss of generality, we may assume that α and α+1 are the smaller components
of v and w respectively. If there is no {α, α − 1}-connection in K, then since v has
degree ≥ 2, there must be a further {α, α + 1}-connection incident on v. Likewise, if
there is no {α + 1, α + 2}-connection, there must be a further {α, α + 1}-connection
incident on w. Otherwise, let x be any vertex of K whose smaller component is < α,
and let y be any vertex of K whose smaller component is > α. Then, since {v, w}
cannot be a bridge, there must be a path in K from x to y that does not use the edge
{v, w}. But if the smaller components of the vertices in this path are listed in order,
they must form a sequence with initial value < α, final value > α, and each pair of
successive values differing by 1. It follows that the sequence must contain α followed
by α + 1, and the corresponding edge of K is a second {α, α + 1}-connection. 2
Lemma 2.5.2 Let K be a kernel of rF (π). If the leftmost occurrence of a right breakpoint (with respect to K) is in position i of π, then position i − 1 is not a breakpoint of
K.
Proof We can assume that π(i − 1) = α and π(i) = β are the components of the
leftmost right breakpoint, v say, of K. Since v is a right breakpoint, there must be
a breakpoint with components α0 and β 0 , say, lying to the left of v in π. Since there
is no right breakpoint in position i − 1, a second {α, α0 }-connection can only arise if
α0 is sandwiched between β 0 and β ∗ , so that all the {α, α0 }-connections involve the
breakpoint v. If there is a breakpoint of K, u say, in position i − 1, it must be a left
breakpoint, so α∗ must lie to the right of α, and any {α, α∗ }-connections must involve
the breakpoint u, since they cannot involve v. But then there cannot be a path in K
from a breakpoint having α0 as a component to one having α∗ as a component, so u
and v cannot both be part of K. Hence, position i − 1 is not a breakpoint of K. 2
Lemma 2.5.3 If α1 and α2 are the smaller (respectively larger) components of two
vertices in a kernel K, then each value between α1 and α2 is also the smaller (respectively larger) component of a vertex in K.
Proof Let α1 and α2 be the smaller components of vertices v and w respectively. Since
the kernel K is connected, there is a path from v to w, and the successive vertices on
this path have smaller components differing by exactly one. Hence the path contains
vertices with smaller components covering all values between α1 and α2 . The argument
for larger components is similar. 2
Lemma 2.5.4 For a permutation π of length n, the breakpoints of a kernel are in
positions i, i + 1, . . . , i + m − 1 and j, j + 1, . . . , j + m − 1 for some i, j and m with 1 ≤ i,
i + m < j, j ≤ n − m + 2. Further, any edge of K connects breakpoints in positions p
and q for some p, q with i ≤ p ≤ i + m − 1 and j ≤ q ≤ j + m − 1.
Chapter 2. Sorting by Reversals
41
Proof Suppose α is one-sided with respect to K. If there is an {α, α + 1}-connection
in K, then there must be at least two such connections, by Lemma 2.5.1. Hence there
can be no {α, α − 1}-connection because there would have to be two such connections,
and this would imply a vertex of degree 4 in K which is impossible. Similarly, if there
is an {α, α − 1}-connection there can be no {α, α + 1}-connection. So, by Lemma 2.5.3,
the only candidates for one-sided elements are
(a) the minimum of the smaller components of breakpoints in K;
(b) the maximum of the smaller components of breakpoints in K;
(c) the minimum of the larger components of breakpoints in K;
(d) the maximum of the larger components of breakpoints in K.
Since there can be at most four one-sided elements it follows that the breakpoints of
K lie in at most two intervals of the permutation. It is a consequence of Lemma 2.5.2
that the breakpoints must lie in two disjoint intervals [i, i + l − 1] and [j, j + m − 1]
with i + l < j, and that all breakpoints in the first interval are left breakpoints and
all those in the second interval are right breakpoints. Finally, since we are assuming
that rF (π) has a perfect matching, it follows that K does also, and since all edges of
K connect breakpoints in the two separate intervals, it follows that l = m. 2
If all the vertices in a kernel have degree 2, then the kernel is just a cycle (and
there are only two possible matchings of the vertices in this kernel). We now seek to
characterise kernels containing at least one vertex of degree 3.
Lemma 2.5.5 If a breakpoint (vertex) v has degree 3 in a kernel K of a permutation
π, then it is adjacent to three successive breakpoints x, y and z of π, and the middle
one of these, y, must be adjacent to the breakpoints u and w on either side of v.
Proof Let the components of v be α and β, with α to the left of β, and suppose,
without loss of generality, that v is a left breakpoint. The fact that v is adjacent to
three successive breakpoints is immediate. There are then two cases to consider:
(a) x, y and z occur in the sequence α0 β 0 α∗ β ∗ ;
(b) x, y and z occur in the sequence β 0 α0 β ∗ α∗ .
We prove the result only for case (a), the proof for case (b) being entirely analogous.
So the permutation takes the form
. . . . . . . . . . . . αβ . . . . . . . . . . . . α0 β 0 α∗ β ∗ . . . . . . . . . . . .
To get a second {α, α0 }-connection, there must be a breakpoint of K immediately to
the left of α, and to enable that breakpoint to have degree 2 in K, the value to the left
of α must be β 00 or β ∗∗ . Similarly, the value to the right of β must be α00 or α∗∗ . So
this leads to four subcases, as follows:
Chapter 2. Sorting by Reversals
case
case
case
case
42
(i) . . . . . . . . . . . . β 00 αβα∗∗ . . . . . . . . . . . . α0 β 0 α∗ β ∗ . . . . . . . . . . . .
(ii) . . . . . . . . . . . . β ∗∗ αβα00 . . . . . . . . . . . . α0 β 0 α∗ β ∗ . . . . . . . . . . . .
(iii) . . . . . . . . . . . . β ∗∗ αβα∗∗ . . . . . . . . . . . . α0 β 0 α∗ β ∗ . . . . . . . . . . . .
(iv) . . . . . . . . . . . . β 00 αβα00 . . . . . . . . . . . . α0 β 0 α∗ β ∗ . . . . . . . . . . . .
We will show that cases (ii), (iii) and (iv) are impossible. The conclusion of the lemma
follows immediately in case (i).
case (ii) To enable breakpoint (β, α00 ) to have degree 2, β ∗ must be followed by α000 . To
enable breakpoint (β ∗ , α000 ) to have degree 2, β ∗∗ must be preceded by α0000 . To enable
breakpoint (β 0 , α∗ ) to have degree 2, α∗∗ must be a component of some breakpoint in
K. But then it is easy to verify that two {α∗ , α∗∗ }-connections cannot exist, showing
that this case cannot arise.
case (iii) To enable breakpoint (β ∗∗ , α) to have degree 2, β ∗∗∗ must precede α0 . To
enable breakpoint (α0 , β 0 ) to have degree 2, (α00 , β 00 ) (or (β 00 , α00 )) must be a breakpoint
in K.
Now, since K contains no bridge there must be a path in K from breakpoint (α 0 , β 0 )
to (α00 , β 00 ) that does not use the edge joining them, and the first step in this path must
lead to (α, β), the only other available edge from (α0 , β 0 ). This path must pass through,
at some point, the other breakpoint with component β 0 — i.e., (β 0 , α∗ ) — so this may
as well be the next step on the path. It must also pass through the other breakpoint
with component α — i.e., (β ∗∗ , α) — but to reach that vertex from (β 0 , α∗ ), the path
must pass through the other breakpoint with component β, namely (β, α ∗∗ ). But then
both breakpoints having β as a component would have already appeared in the path,
and so it would impossible to get from (β ∗∗ , α) to (α00 , β 00 ). Hence this case cannot
arise.
case (iv) To enable breakpoint (β, α00 ) to have degree 2, β ∗ must be followed by α000 .
To enable breakpoint (α∗ , β ∗ ) to have degree 2, there must be a breakpoint with components α∗∗ and β ∗∗ . This case can now be ruled out by an argument analogous to
that used for case (iii), considering the possibility of an alternative path from (α ∗ , β ∗ )
to (α∗∗ , β ∗∗ ). 2
Lemma 2.5.6 If any vertex in a kernel has degree 3, then for some i < j, the edges
in the kernel are either
(i) {vi , vj }, {vi , vj+1 }; {vi+m−1 , vj+m−2 }, {vi+m−1 , vj+m−1 }; and
{vk , vk+j−i−1 }, {vk , vk+j−i }, {vk , vk+j−i+1 } for i + 1 ≤ k ≤ i + m − 2; or
(ii) {vi , vj+m−1 }, {vi , vj+m−2 }; {vi+m−1 , vj+1 }, {vi+m−1 , vj }; and
{vk , vi+j+m−k−2 }, {vk , vi+j+m−k−1 }, {vk , vi+j+m−k } for i+1 ≤ k ≤ i+m−
2.
Proof Suppose vk has degree 3 for some k (i ≤ k ≤ i + m − 1). Then vk is adjacent
in K to vl−1 , vl , and vl+1 for some l (j < l < j + m − 1). By Lemma 2.5.5, it follows
43
Chapter 2. Sorting by Reversals
07
72
29
96
61
18
83
3 10
10 5
4 11
Figure 2.18: rF ∗ (π3 )
that i < k < i + m − 1 and that positions k − 1, k, k + 1, k + 2 and l − 1, l, l + 1, l + 2
are occupied in one of the two possible ways shown:
case (a) . . . . . . . . . β 00 αβα∗∗ . . . . . . . . . α0 β 0 α∗ β ∗ . . . . . . . . .
case (b) . . . . . . . . . β ∗∗ αβα00 . . . . . . . . . β 0 α0 β ∗ α∗ . . . . . . . . ..
If m = 3 then we see that these two possibilities correspond to cases (i) and (ii) in
the statement of the lemma. Otherwise, one of vk+1 or vk−1 has degree 3. Applying
Lemma 2.5.5 leads to configurations (i) and (ii) with m = 4. The argument may be
continued inductively to complete the proof. 2
Let us call kernels of the form described in Lemma 2.5.6 parts (i) and (ii) overlapped
and nested kernels, respectively. A kernel in which every vertex has degree 2 will be
called a cyclic kernel.
At this point it may be helpful to include some illustrative examples.
Example
π3 = [0 7 2 9 6 1 8 3 10 5 4 11].
The permutation π3 contains an overlapped kernel with 6 breakpoints (see Figure 2.18).
It is a split kernel — see the definition below, split by the breakpoint (9, 6), since it is
forced to be matched in an oriented cycle with the breakpoint (10, 5). 2
Example
π4 = [0 11 2 9 4 7 6 5 8 3 10 1 12].
The permutation π4 contains a nested kernel that includes all 10 breakpoints (see
Figure 2.19). It is a whole kernel — see the definition below. 2
Example
π5 = [0 8 4 10 2 6 11 3 7 1 9 5 12].
The permutation π5 contains a cyclic kernel with 10 breakpoints (see Figure 2.20). It
is a split kernel, split by the single breakpoint (6, 11). 2
Now that we have characterised the structure of kernels, we have to consider how
to deal with them when they arise. Recall that a permutation is reversal-tight if and
44
Chapter 2. Sorting by Reversals
0 11
11 2
29
94
47
58
83
3 10
71
19
10 1
1 12
95
5 12
Figure 2.19: rF ∗ (π4 )
08
84
4 10
10 2
26
6 11
11 3
37
Figure 2.20: rF ∗ (π5 )
Chapter 2. Sorting by Reversals
45
only if there is a perfect matching M in the graph rF (π) such that every component of
rR(CM ) is oriented. For a given kernel K, and a perfect matching M in rF (π), denote
by KM the subgraph of rR(CM ) comprising of those components that contain vertices
arising from 2-cycles (of rG(π)) represented by edges in K and M .
According to Lemma 2.5.4, a kernel K has its left and right breakpoints in two
disjoint intervals of the permutation. Suppose that there is a 2-reversal with exactly
one of its endpoints lying between these two intervals, and that this 2-reversal is either
part of another kernel or is forced, during the construction of rF ∗ (π) from rF (π), to
be part of any perfect matching. In this case, K will be called a split kernel. If there
is no such 2-reversal the kernel will be called whole. If rF (π) has a perfect matching
M such that every component of KM is oriented, then K will be called a reversal-tight
kernel.
Lemma 2.5.7 Let π be a permutation with a kernel K, and suppose that the graph
rF (π) has a perfect matching.
(i) If K is a whole kernel which is overlapped or nested then it is reversaltight.
(ii) If K is a whole kernel which is cyclic, then there is a polynomial-time
algorithm to check whether it is reversal-tight.
(iii) If K is a split kernel then it is reversal-tight.
Proof We prove the three cases separately:
(i) Let K be an overlapped kernel, with breakpoints vi , . . . , vi+m−1 and vj , . . . ,
vj+m−1 as in the statement of Lemma 2.5.6. If we take a perfect matching in K that
includes the edges {vi , vj+1 } and {vi+1 , vj }, and it is clear that such a matching exists,
then both of these edges represent exposed 2-reversals, and as vertices in K M , both are
represented by vertices that are adjacent to all of the vertices arising from K. Hence
the single component of KM is oriented.
On the other hand, if K is a nested kernel, we may take the perfect matching that
pairs vi with vj+m−1 , vi+m−1 with vj , and, for k = 1, . . . , m − 2, vi+k with vj+m−1−k .
Every 2-reversal in this perfect matching is exposed, so that every component of K M
is oriented.
(ii) In this case, since the kernel is whole, the components of KM contain reversals only
on breakpoints of K. There are only two possible perfect matchings in K, and it is a
simple matter to check, in polynomial time, whether for either one, all the components
of KM are oriented.
(iii) For any perfect matching M of rF (π), KM will be a connected component, since
the edge that splits the kernel (or any of the chosen edges of a splitting kernel) will be
represented in KM by a vertex that is adjacent to all vertices arising from K. So in
this case, it suffices to determine whether K contains a perfect matching that includes
at least one exposed 2-reversal.
Chapter 2. Sorting by Reversals
46
If K is cyclic (necessarily with an even number of vertices) then there are just
two perfect matchings in K. The leftmost breakpoint in K is an end-point of two
2-reversals, one of which is exposed and one of which is hidden. So we may choose the
matching that includes this particular exposed 2-reversal.
If K is overlapped or nested then the perfect matchings described in the proof of
(i) each include at least one exposed 2-reversal, and this is all that is required. 2
Note that, by the definition of a kernel, the choices we make for the matching
restricted to each kernel are independent. So we can construct a global matching from
the matchings for each kernel. Therefore, it follows from Lemma 2.5.7 that, if π is a
permutation for which the graph rF (π) contains a perfect matching, then in order to
determine whether π is reversal-tight, the following steps suffice:
1. identify the kernels;
2. for any whole cyclic kernel K, establish whether at least one of the two possible
+
matchings gives components of rR(CM
) that are all oriented;
+
3. establish whether all components of rR(CM
) that do not involve kernels are all
oriented.
We have therefore proved the following theorem:
Theorem 2.5.2 For a given permutation π, it can be established in polynomial time
whether or not π is reversal-tight.
This result has been obtained independently by Tran [Tra97]. He proves the first
half (Theorem 2.5.1) of this result in roughly the same way as we do. However, he
proves the second half in quite a different way from us. In effect, Tran shows that if π
+
is reversal tight then all components of rR(CM
) are oriented, where M is a maximum
weight perfect matching of rF (π) with edges given weight +1 if they represent an
exposed 2-reversal and weight −1 otherwise. So his proof of this part is simpler than
our proof because it does not need to describe the complicated structure of kernels.
2.6
Conclusion and open problems
The 3/2-approximation algorithm of Section 2.4 has the best known approximation
guarantee for sorting by reversals. It is an open problem to find a polynomial-time
algorithm with a better approximation guarantee.
The question of whether sorting by reversals is MAXSNP-complete [Pap94], and
the related question of whether there is a polynomial time approximation scheme for
the problem, are both open.
Chapter 3
Sorting by Transpositions
3.1
Introduction
In this chapter we study the problem of sorting by transpositions. The computational
complexity of this problem remains open, but we make a number of contributions to
the better understanding of this problem. The significant new results of this chapter
are as follows:
• we analyse the structure of cycles in the so-called transposition cycle graph,
leading to a better understanding of how these cycles affect the problem (Section
3.3). This includes defining cycles called knots, and describing a method for
dealing with such cycles in an efficient manner.
• we present a new 3/2-approximation algorithm for sorting by transpositions that
is somewhat simpler than that presented by Bafna and Pevzner (Section 3.4).
• we present an improved lower bound for sorting by transpositions (Section 3.5).
• we present empirical evidence that in practice we can solve most instances of
sorting by transpositions (Section 3.6).
Some other interesting new results in this chapter include:
• a proof that it is never necessary to break apart any strips in a minimal length
sequence of transpositions that sorts a permutation (Section 3.2.2).
• a description of a family of permutations whose transposition distance can be
arbitrarily far from Bafna and Pevzner’s lower bound for transposition distance
(Section 3.5.2).
• a proof that the transposition distance of the reverse permutation R n is bn/2c + 1
(Section 3.5.3).
We begin the chapter with a section containing fundamental definitions, lemmas
and theorems used to study the problem of sorting by transpositions.
47
Chapter 3. Sorting by Transpositions
3.2
3.2.1
48
Fundamental definitions and results
Definitions
Let π be a permutation of length n. The transposition τ = τ (i, j, k) (where 1 ≤
i < j < k ≤ n + 1) transforms π into π · τ = [π(0) . . . π(i − 1) π(j) . . . π(k −
1) π(i) . . . π(j − 1) π(k) . . . π(n + 1)]. The transposition distance, td(π), between π
and ı is the length of a shortest sequence of transpositions that transforms π into ı. A
sequence of transpositions that transforms π into ı is called a sorting sequence. Sorting
by transpositions is the problem of finding a sorting sequence for π of length td(π).
A transposition breakpoint is a position i in the permutation such that 1 ≤ i ≤ n+1
and π(i) − π(i − 1) 6= 1. The number of transposition breakpoints in a permutation π
is represented by tb(π). By contrast, a transposition adjacency is a position i in the
permutation such that 1 ≤ i ≤ n+1 and π(i)−π(i−1) = 1. A strip is a substring π[i..j]
of π (i < j) such that i and j + 1 are breakpoints and there are no breakpoints between
these positions. Note that, for conciseness, in the rest of this chapter we may talk of
breakpoints and adjacencies where, of course, we mean transposition breakpoints and
transposition adjacencies.
Let f (π) be some function on π, e.g., tb(π). We are often interested in the change
in value of f (π) as a result of applying a transposition τ . We shall use ∆f (π, τ ) to
denote the change. Formally, ∆f (π, τ ) = f (π · τ ) − f (π).
It is easy to show that the value of ∆tb(π, τ ) is characterised by the following
lemma.
Lemma 3.2.1 For all permutations π and transpositions τ , ∆tb(π, τ ) ≥ −3.
This lemma leads to the following lower bound for transpositions distance, since
the identity permutation contains no breakpoints.
Theorem 3.2.1 For all permutations π, td(π) ≥ dtb(π)/3e.
3.2.2
Transposition equivalent permutations
A little thought suggests that it is unlikely that a shortest sorting sequence would ever
have to break apart a strip, since that strip would later have to be put back together.
In fact this intuition is correct, as we prove below.
Let r be the number of strips in π, excluding any initial strip beginning with 1
and any final strip ending with n. We define the minimal permutation gl(π) to be the
permutation of {1, . . . , r} formed from π by ‘gluing’ all the adjacencies together, i.e.,
replacing each strip by a single element. Note that if π begins with 1 then the strip at
the start of π is removed completely in gl(π) and, similarly, if π ends with n then the
strip at the end of π is removed completely in gl(π).
Example If π = [4 5 6 3 1 2 7] then gl(π) = [3 2 1]. 2
49
Chapter 3. Sorting by Transpositions
0
3
4
1
2
5
Figure 3.1: tG(π), for π = [3 4 1 2].
Theorem 3.2.2 For every permutation π, td(π) = td(gl(π)).
Proof Clearly td(π) ≤ td(gl(π)), since any transposition on gl(π) may be mimicked
by a transposition on π.
We now show that td(gl(π)) ≤ td(π). Let π be a permutation of length n. We
may assume that gl(π) is length m, such that m ≤ n. Form a string Sπ of length
n by inserting n − m asterisks (∗) into gl(π), such that Sπ (i) is an asterisk if i is an
adjacency in π, or π(j) = j for all j ≥ i. For example, if π = [1 3 4 5 7 8 2 6 9], then
gl(π) = [2 4 1 3], and Sπ = [∗ 2 ∗ ∗ 4 ∗ 1 3 ∗]. Then any transposition on π can be applied
to Sπ , and ignoring asterisks this transposition can be applied to gl(π). Note that if any
transposition on Sπ moves a block that consists only of asterisks then the corresponding
transposition on gl(π) does nothing, so can be ignored. A sequence of transpositions
that sorts π will sort Sπ (ignoring asterisks), and so a sequence of transpositions of at
most the same length exists that sorts gl(π). Hence td(gl(π)) ≤ td(π), and the theorem
has been proved. 2
We say that permutations π and φ are transposition equivalent if gl(π) = gl(φ).
As a consequence of Theorem 3.2.2 we can essentially restrict our attention to
permutations with no adjacencies. A strip free permutation π is a permutation of
length n such that π contains no adjacencies, π(1) 6= 1, and π(n) 6= n.
3.2.3
The transposition cycle graph
The transposition cycle graph of π, tG(π), is a directed edge-coloured graph with vertex
set {0, . . . , n + 1}, grey edge set {(i, i + 1) : 0 ≤ i ≤ n}, and black edge set {(π(i), π(i −
1)) : 1 ≤ i ≤ n + 1}. Note that the edge (u, v) is directed from u to v. This graph was
introduced by Bafna and Pevzner [BP95b].
Example Let π be the permutation [3 4 1 2]. Then tG(π) is shown in Figure 3.1.
Note that by convention tG(π) is drawn without any circles representing the vertices.
We also, by a convention to be explained later, draw tG(π) so that the black edge
(π(i), π(i − 1)) points directly to the grey edge (π(i − 1), π(i − 1) + 1), and the grey
edge (i, i + 1) points directly to the black edge (i + 1, π(π −1 (i + 1) − 1)). 2
Chapter 3. Sorting by Transpositions
50
The black edge (π(i), π(i − 1)) is denoted by bi . This edge, bi , is also known as the
black edge at position i. As shown in Figure 3.1, we draw tG(π) so that the black edges
occur in the order b1 , . . . , bn+1 from left to right. It is because of this convention that
we define an edge bi to be to the left of bj if i < j. Similarly, we define bi to be to the
right of bj if i > j. The leftmost edge of a set of black edges is the edge bi from the
set such that i is minimised. Similarly, the rightmost edge of a set of black edges is
the edge bi from the set such that i is maximised. A grey edge (i, i + 1) is said to be
directed to the right if π −1 (i) < π −1 (i + 1). Similarly, a grey edge is said to be directed
to the left if π −1 (i) > π −1 (i + 1). In Figure 3.1, the grey edge (2, 3) is directed to the
left, and all the other grey edges are directed to the right.
A path in tG(π) is an alternating path if its edges alternate in colour. Similarly, a
cycle is an alternating cycle if its edges alternate in colour.
At each vertex of tG(π), each incoming edge can be paired with an outgoing edge
of the alternative colour. This pairing of edges decomposes the graph into alternating
cycles, and furthermore the decomposition is unique, since each vertex has at most
one incoming edge of each colour, and one outgoing edge of each colour. Such cycle
decompositions are obvious in our figures because of the conventions we use when
drawing cycle graphs. The number of alternating cycles in the cycle decomposition of
tG(π) is denoted by tc(π). The maximum value of tc(π) is n + 1, and is achieved only
when π is the identity permutation. In the rest of this chapter it is assumed that all
the cycles we consider are alternating cycles.
Let C be a cycle in tG(π). The length of C, l(C), is the number of black edges in
C. A k-cycle is a cycle of length k. A proper cycle is a cycle that has length greater
than one. A long cycle is a cycle that has length greater than two. If l(C) is odd then
the cycle is odd, and otherwise the cycle is even. The number of odd cycles in tG(π) is
denoted by tcodd (π). Similarly, tceven (π) denotes the number of even cycles in tG(π).
Example If π is the permutation [3 4 1 2] then tc(π) = tcodd (π) = 3 and tceven (π) =
0, as illustrated in Figure 3.1. 2
Bafna and Pevzner [BP95b] proved the following lemmas about the number of cycles
and odd cycles in the cycle graph.
Lemma 3.2.2 For all permutations π and transpositions τ , ∆tc(π, τ ) ∈ {−2, 0, 2}.
Lemma 3.2.3 For all permutations π and transpositions τ , ∆tcodd (π, τ ) ∈ {−2, 0, 2}.
A transposition τ is a t-transposition if ∆tcodd (π, τ ) = t. Lemma 3.2.3, combined
with the fact that the identity permutation contains n + 1 odd length cycles, allowed
Bafna and Pevzner to prove the following lower bound for transposition distance.
Theorem 3.2.3 For all permutations π, td(π) ≥ (n + 1 − tcodd (π))/2.
51
Chapter 3. Sorting by Transpositions
0
6
1
5
7
12
4
11
2
9
3
8
10
13
Figure 3.2: tG(π), for π = [6 1 5 7 12 4 11 2 9 3 8 10].
The positions of the leftmost black edge and the rightmost black edge of a cycle C
are denoted by Cmin and Cmax respectively. More formally, these values can be defined
as Cmin = min{i : bi is an edge in C}, and Cmax = max{i : bi is an edge in C}.
By convention, a cycle is described by listing the positions of the black edges of
the cycle in the order they occur in the cycle, starting from position Cmax . A cycle
C that visits black edges in the order bi1 , . . . , bil(C) is denoted by (i1 , . . . , il(C) ). (Note
i1 = Cmax .)
Example Let π be the permutation [6 1 5 7 12 4 11 2 9 3 8 10]. The cycle graph
of this permutation contains three cycles: (4, 1, 2), (12, 9, 10), and (13, 7, 3, 8, 5, 11, 6),
as shown in Figure 3.2. 2
The transposition τ = τ (i, j, k) is said to act on black edges bi , bj and bk of tG(π),
since the transformation from tG(π) to tG(π · τ ) involves removing these three edges
and introducing three new edges, whereas all the other grey and black edges of tG(π)
remain intact in tG(π · τ ).
Three black edges of tG(π) form an oriented triple if τ , the transposition that acts
on them, is such that ∆tc(π, τ ) = 2. Three black edges of a cycle form an unoriented
triple if they do not form an oriented triple.
Example In Figure 3.2, (b1 , b2 , b4 ) is an oriented triple and (b3 , b5 , b8 ) is an unoriented triple. 2
An oriented cycle contains an oriented triple of black edges. A cycle that is not
oriented is unoriented. Bafna and Pevzner noted the following theorem characterising
unoriented cycles.
Theorem 3.2.4 Cycle C is unoriented if and only if (i1 , . . . , il(C) ) is a decreasing
sequence.
When a transposition is applied to a permutation the cycle graph of the resulting
permutation is the same as the original cycle graph except that three black edges
have been removed, and three black edges have been introduced. It is likely that the
positional labelling (bi ) of a black edge that was not removed or introduced will have
changed as the result of the transposition. We often want a label for an edge that does
Chapter 3. Sorting by Transpositions
52
not change if the edge is not removed. We use the letters s, t, u, v, w, x, y and z
as positionally independent labels of black edges. The position of such a black edge u
in π is denoted by p(π, u). For brevity we sometimes write a positionally independent
label, where we should really write the position of the label. For example we may write
u < v in place of p(π, u) < p(π, v).
3.2.4
Even length cycles
Bafna and Pevzner do not give much emphasis to the following results about even
length cycles, but we will show them to be very useful in later sections. The first result
is a lemma that can be derived easily from Lemma 3.2.2 and Lemma 3.2.3.
Lemma 3.2.4 For all permutations π and transpositions τ , ∆tceven (π, τ ) ∈ {2, 0, −2}.
This lemma leads naturally to the following corollary.
Corollary 3.2.1 tceven (π) is an even number, for all permutations π.
Proof This follows from the preceding lemma and the fact that the cycle graph of the
identity permutation contains no even length cycles. 2
This corollary allows us to prove the following useful lemma, that helps us determine
if it is possible to apply a 2-transposition to a permutation.
Lemma 3.2.5 For all permutations π, if tG(π) contains an even cycle we may apply
a 2-transposition on π.
Proof By Corollary 3.2.1, if tG(π) contains an even cycle then it must contain at least
two even cycles. Let C and D be even cycles of tG(π). Let x and y be black edges
of C such that the alternating path connecting x to y (but excluding both edges) in
tG(π) contains an even number of black edges. Let z be any black edge of D. Then
the transposition acting on black edges x, y and z is a transposition that increases the
number of odd length cycles in the cycle graph by 2 (although it does not increase the
total number of cycles in the cycle decomposition). 2
3.3
Fully oriented cycles
In this section we describe how to determine if a 2-transposition exists on a permutation. In particular, we achieve this by characterising those cycles (the fully oriented
cycles) that admit 2-transpositions that act on their edges. We also describe a method
for efficiently dealing with cycles that are oriented, but do not admit 2-transpositions
on their edges. We then compare our fully oriented cycles with Bafna and Pevzner’s
strongly oriented cycles. Finally we define cycles, super oriented cycles, that permit
a sequence of 2-transpositions on their edges, and show a preliminary result towards
characterising these cycles.
Chapter 3. Sorting by Transpositions
3.3.1
53
Fully oriented triples
Three black edges x, y and z of tG(π) form a fully oriented triple (x, y, z) if τ , the
transposition on those edges, is a 2-transposition, and x, y, and z are all part of the
same cycle in tG(π). A cycle that contains a fully oriented triple is said to be fully
oriented.
Fully oriented triples are useful because of the following simple proposition.
Proposition 3.3.1 Let τ be a 2-transposition that acts on black edges x, y, and z of
tG(π). Then either x, y and z are all part of the same cycle in tG(π), or they are from
two even length cycles in tG(π)
Proof Edges x, y, and z are edges of one, two or three cycles in tG(π). All possible
cases are shown in Figure 3.3. Note that vertices are not shown in this figure because
the figure represents only part of tG(π) and so there may be many vertices between
the black edges. In this figure dotted grey edges represent alternating paths in tG(π)
that start and finish with a grey edge.
A transposition can only be a 2-transposition if it is like that shown in Figure 3.3
(b), (c), (d) or (e). So x, y, and z must be part of one or two cycles. 2
As a consequence of this proposition, a 2-transposition either acts on two even
length cycles, or it acts on a fully oriented triple of edges. Now, if a 2-transposition
acts on two even cycles, then we can show that the edges the transposition acts on must
be arranged as described in the proof of Lemma 3.2.5. We characterise 2-transpositions
on one cycle by characterising fully oriented triples and fully oriented cycles.
Given a cycle we can easily tell, using Theorem 3.2.4, whether the cycle is oriented,
or unoriented. However an oriented cycle is not necessarily a fully oriented cycle. For
example, if π = [4 3 2 1] then the cycle (5, 3, 1, 4, 2) in tG(π) (Figure 3.4) is oriented,
but not fully oriented. What conditions make an oriented cycle fully oriented? We
answer this question in Section 3.3.3, but to help us find the answer we need a new
notation for cycles, that we call the canonical labelling of a cycle.
3.3.2
Canonical labellings
With the cycle notation introduced in Section 3.2.3 we can tell if a cycle is oriented, but
the notation is not very useful for anything else, in particular it is difficult to compare
the structure of two cycles using this notation.
We now introduce a new cycle notation called the canonical labelling in which the
structures of different cycles can be compared more easily. The first step towards
obtaining this new notation for a cycle C is to (re)label its black edges 1, 2, . . . , l(C)
so as to preserve the relative positional order of the edges. So edge b Cmin is labelled 1,
the edge bi of C such that i is minimised but greater than Cmin is labelled 2, and so
on up to bCmax , which is labelled l(C).
54
Chapter 3. Sorting by Transpositions
(a)
(b)
(c)
(d)
(e)
(f)
Figure 3.3: How a transposition changes the cycle graph.
0
4
3
2
1
Figure 3.4: tG(π), for π = [4 3 2 1].
5
Chapter 3. Sorting by Transpositions
55
The canonical labelling, L(C), of the cycle C is the permutation of the set {1, . . . ,
l(C)} obtained from this labelling by starting with the edge labelled 1 then reading
the labels in the order they occur around the cycle. Note that all canonical labellings
begin with 1.
It will prove useful to split the canonical labelling into two subsequences. The
outer sequence O(C) is defined by taking the 1st, 3rd, etc. labels from the canonical
labelling. The inner sequence I(C) is defined by taking the 2nd, 4th, etc. labels from
the canonical labelling. The kth elements of these sequences are denoted by i(C, k)
and o(C, k) respectively.
Example Let π be the permutation [6 1 5 7 12 4 11 2 9 3 8 10]. The cycle graph
of this permutation contains three cycles: (4, 1, 2), (12, 9, 10), and (13, 7, 3, 8, 5, 11, 6),
as shown in Figure 3.2. These cycles have canonical labellings [1 2 3], [1 2 3], and
[1 5 2 6 3 7 4]. The third cycle has outer labelling [1 2 3 4] and inner labelling [5 6 7].
From this new notation it is perhaps more obvious that two of the cycles are very
similar. 2
In the old cycle notation, a cycle is unoriented if it is represented by a decreasing
sequence. So we may restate Theorem 3.2.4 in terms of canonical labellings as follows
Theorem 3.3.1 A cycle C is unoriented if and only if
L(C) = [1 l(C) l(C) − 1 . . . 2].
3.3.3
Fully oriented cycles
We now show how to determine if a cycle is fully oriented based on its canonical
labelling. We show that all even oriented cycles are fully oriented, whereas some
odd oriented cycles are not fully oriented. First we need a proposition characterising
oriented triples.
Proposition 3.3.2 Let [x y z] be a subsequence of the canonical labelling of a cycle.
Then (x, y, z) is an oriented triple if and only if x < y < z, or y < z < x, or z < x < y.
Proof If x, y and z satisfy one of these three conditions, then the cycle must be
oriented. The transposition on edges x, y, and z splits the cycle into three cycles, as
shown in Figure 3.3 (b). If x, y and z do not satisfy any of the three conditions, then
they must be arranged as shown in Figure 3.3 (a), and the transposition, as shown,
does not increase the number of cycles. Therefore the edges form an unoriented triple.
2
Now we show that every oriented even length cycle is a fully oriented cycle.
Lemma 3.3.1 Let [x y z] be a subsequence of the canonical labelling of an even length
cycle C, such that x < y < z, or y < z < x, or z < x < y. Then (x, y, z) is a fully
oriented triple if and only if [x y z] is neither a subsequence of I(C) nor a subsequence
of O(C).
56
Chapter 3. Sorting by Transpositions
Proof A transposition on these three edges will split C into 3 cycles (Proposition
3.3.2). If [x y z] is a subsequence of I(C) or O(C) then the paths between x, y and z
all contain an odd number of black edges. So the resulting cycles are all even in length.
However if [x y z] is not a subsequence of I(C) or O(C) then two of the resulting cycles
will be odd in length. 2
Theorem 3.3.2 If C is an even length cycle that is oriented, then C is also fully
oriented.
Proof Since C is oriented it must have the canonical labelling [1 . . . i . . . j . . .] such
that i < j. Obviously, 1 is in O(C). If at least one of i and j is in I(C) then (1, i, j) is
a fully oriented triple, by Lemma 3.3.1. If both of i and j are in O(C) then there must
be a k in I(C) such that C has canonical labelling [1 . . . i . . . k . . . j . . .]. If k < j
then (1, k, j) is a fully oriented triple. If k > j then (1, i, k) is a fully oriented triple. 2
We now show that no analogous result can be proved for odd length cycles. However,
we are able to characterise those odd length cycles that are oriented but not fully
oriented.
Lemma 3.3.2 Let [x y z] be a subsequence of the canonical labelling of an odd length
cycle C. Then (x, y, z) is a fully oriented triple if and only if one of the following six
conditions on x, y and z is true:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
x < y < z,
x < y < z,
y < z < x,
y < z < x,
z < x < y,
z < x < y,
x
x
x
x
x
x
and
and
and
and
and
and
z
z
z
z
z
z
are
are
are
are
are
are
in
in
in
in
in
in
O(C) and y is in I(C)
I(C) and y is in O(C)
O(C) and y is in I(C)
I(C) and y is in O(C)
O(C) and y is in I(C)
I(C) and y is in O(C)
Proof Suppose that x, y and z satisfy one of these conditions. Then the three edges
must form an oriented triple (Proposition 3.3.2). Furthermore the paths in C between
x, y and z all contain an even number of black edges, and so the transposition on these
edges increases the number of odd length cycles in the cycle graph by two. So the edges
form a fully oriented triple.
Suppose that x, y and z do not satisfy any of the above conditions. If (x, y, z) is
not an oriented triple then it cannot be a fully oriented triple. If (x, y, z) is an oriented
triple then two of the paths in C between x, y and z contain an odd number of black
edges, and so the transposition on these edges increases the number of even length
cycles in the cycle graph. So in neither case is (x, y, z) a fully oriented triple 2
Theorem 3.3.3 For an odd length cycle of length 2r+1 (r > 1), the canonical labelling
[1 r + 2 2 r + 3 3 . . . 2r + 1 r + 1] is the only canonical labelling possible for a cycle
that is oriented, but not fully oriented.
Chapter 3. Sorting by Transpositions
57
Proof Clearly this is a canonical labelling of an oriented cycle. However the cycle
is not fully oriented, because O(C) is the sequence [1, 2, . . . , r + 1] and I(C) is the
sequence [r + 2, r + 3, . . . , 2r + 1]. Hence, it is impossible to find edges x, y and z that
satisfy any of the six conditions of Lemma 3.3.2.
We now show that this is the only possible canonical labelling for an oriented cycle
that is not fully oriented. Let C be an odd length cycle of length 2r +1 that is oriented,
but is not fully oriented.
In particular let us first suppose that o(C, k) ≤ r + 1 for all values of k. This
restriction means that o(C, k) < i(C, l) for all values of k and l. If i(C, k) > i(C, l) but
k < l then by Lemma 3.3.2 the black edges at positions o(C, l), i(C, l), and i(C, k) form
a fully oriented triple. Therefore I(C) must be the sequence [r + 2 r + 3 . . . 2r + 1].
Similarly O(C) must be the sequence [1 2 . . . r + 1].
This means that the canonical labelling [1 r + 2 2 r + 3 3 . . . 2r + 1 r + 1] is the
only canonical labelling that represents a cycle that is not fully oriented and for which
o(C, k) ≤ r + 1 for all values of k. So in any alternative canonical labelling of a cycle
that is not fully oriented there must be an o(C, k) such that o(C, k) > r + 1. Let C be
a cycle with just such an alternative canonical labelling.
Let o(C, k) be the largest value in O(C), and let i(C, l) be the smallest value in
I(C). Note that o(C, k) > r + 1 and so i(C, l) ≤ r + 1. In particular l ≥ k because
otherwise (1, i(C, l), o(C, k)) would be a fully oriented triple, by Lemma 3.3.2. So C
has canonical labelling [1 . . . o(C, k) . . . i(C, l) . . .].
Now since C is not fully oriented it must be the case that i(C, m) > o(C, k) for all
1 ≤ m < k, since otherwise (i(C, m), o(C, k), i(C, l)) would be a fully oriented triple.
Similarly o(C, m) < i(C, l) for all l < m ≤ r + 1, because otherwise (1, i(C, l), o(C, m))
would be a fully oriented triple. Furthermore o(C, s) > i(C, l) for all 2 ≤ s ≤ l since
otherwise (i(C, 1), o(C, s), i(C, l)) would form a fully oriented triple.
If 2r + 1 were in O(C) then (1, i(C, 1), 2r + 1) would be a fully oriented triple,
so 2r + 1 is in I(C). Suppose i(C, 1) 6= 2r + 1 then i(C, p) = 2r + 1 for some p,
1 < p ≤ r. Then (o(C, 2), i(C, p), o(C, r + 1)) would be a fully oriented triple because
then o(C, r + 1) < i(C, l) < o(C, 2) < i(C, p). So i(C, 1) = 2r + 1. If 2r were in I(C)
then (2r + 1, o(C, 2), 2r) would be a fully oriented triple, so 2r is in O(C). In particular
o(C, k) = 2r, and so k = 2 since i(C, m) > o(C, k) for all 1 ≤ m < k.
Now suppose that C 6= [1 2r + 1 2r . . . 3 2]. Then there must be an i(C, q) <
o(C, r) such that q < r or an o(C, q) < i(C, r) such that q ≤ r and q > 1. If the
first case were true then (2r, i(C, q), o(C, r)) would be a fully oriented triple. If the
second case were true then (2r + 1, o(C, q), i(C, r)) would be a fully oriented triple. So
C = [1 2r + 1 2r . . . 3 2], which is not oriented. 2
We can use these results about the canonical labellings of fully oriented cycles, to
prove that small cycles are less likely to be fully oriented.
Lemma 3.3.3 A cycle of length l is less likely to be fully oriented than a cycle of length
l + 1, assuming all canonical labellings are equally likely.
Chapter 3. Sorting by Transpositions
58
Proof If an odd length cycle (of length 2r + 1) is not fully oriented then it must have
canonical labelling [1 2r + 1 . . . 2], or [1 r + 2 2 r + 3 3 . . . 2r + 1 r + 1]. Now there
are (2r)! possible canonical labellings of an odd length cycle. So the chance that an
odd length cycle is not fully oriented is only 2/(2r)!.
If an even length cycle (of length 2r) is not fully oriented then it must have canonical
labelling [1 2r . . . 2]. So the chance that an even length cycle is not fully oriented is
only 1/(2r − 1)!.
The proof is completed by noting that 1/(2r − 1)! > 2/(2r)! > 1/(2r + 1)!. 2
We shall use this lemma as the basis of a successful heuristic for solving instances
of sorting by transpositions in a later section.
3.3.4
Knots
In this section we prove some results about cycles that are oriented, but not fully
oriented. We call such cycles knots. We define ωr to be the permutation of length 2r
whose cycle graph consists of a knot of length 2r + 1, if such a permutation exists.
Lemma 3.3.4 If r ≡ 1 (mod 3) then ωr is not defined, but ωr exists (and is unique)
for all other values of r.
Proof By Theorem 3.3.3, a knot of length 2r + 1 has canonical labelling [1 r + 2 2 r +
3 3 . . . 2r + 1 r + 1]. So if ωr exists then tG(π) contains r grey edges directed to the
right, namely {(π(i), π(i + r + 2)) : 0 ≤ i ≤ r − 1}, and r + 1 grey edges directed to the
left, namely {(π(i), π(i − (r − 1))) : r ≤ i ≤ 2r}. Therefore all grey edges are directed
from π(i) to π((i + r + 2) mod (2r + 1)) (except the edge (π(r − 1),π(2r + 1))).
Let S be the sequence that is defined by the recurrence relation:
S(0) = 0
S(i + 1) = (S(i) + r + 2) mod (2r + 1), (for i ≥ 0).
Then S = [0 r + 2 3 . . . ]. Suppose that ωr exists. Let us follow grey edges in tG(ωr ),
starting from 0, i.e., visit vertices of tG(π) in the order 0, 1, 2, etc.. Now the positions
of these elements in π is given by the sequence S. Therefore ωr exists if and only if the
period of S is 2r + 1.
Now the period of S is 2r + 1 if and only if r + 2 and 2r + 1 are co-prime, i.e.,
gcd(r + 2, 2r + 1) = 1 where gcd(x, y) is the greatest common divisor of x and y. Now
gcd(r + 2, 2r + 1) = 1 ⇔ gcd(r + 2, r − 1) = 1 ⇔ gcd(3, r − 1) = 1. Therefore ω r exists
if and only if r 6≡ 1 (mod 3). 2
By considering S, the sequence defined in the proof of Lemma 3.3.4, it is easy
to verify that ωr (i) = (i.4(r + 1)/3) mod (2r + 1) if r ≡ 2 (mod 3), and ωr (i) =
(i.(2r/3 + 1)) mod (2r + 1) if r ≡ 0 (mod 3)
Chapter 3. Sorting by Transpositions
59
Figure 3.5: How to sort a knot of length 5.
Lemma 3.3.5 If tG(π) contains a knot, then we may apply a (0, 2, 2)-sequence of
transpositions on the knot.
Proof We can sort ω2 using a (0, 2, 2)-sequence as shown in Figure 3.5. Edges 1, 2,
r + 1, r + 2, and 2r + 1 of a larger knot are connected in exactly the same way as the
five edges of ω2 are connected, except that there is an alternating path (that contains
an even number of black edges) between edges 2 and 2r + 1 instead of a single grey
edge. The sequence of transpositions on ω2 may be applied to these five edges of the
larger knot. The resulting transpositions form a (0, 2, 2)-sequence. 2
Lemma 3.3.5 is sufficiently powerful to obtain the 3/2-approximation algorithm
in Section 3.4. However, below we show that in fact a much stronger result can be
proved, though we warn the reader that the proof of this theorem involves a lengthy
and detailed case by case analysis.
Theorem 3.3.4 If r 6≡ 1
(mod 3) then td(ωr ) = r + 1.
Proof Since, ωr contains only one cycle, and it is not fully oriented, an initial 2transposition cannot be applied to ωr . So td(ωr ) ≥ r + 1.
60
Chapter 3. Sorting by Transpositions
8 5 2 10 7 4 1 9 6 3
1 8 5 2 10 7 4 9 6 3
1 2 10 7 4 9 8 5 6 3
1 2 10 7 8 5 6 3 4 9
1 2 7 8 5 6 3 4 9 10
1 2 5 6 7 8 3 4 9 10
1 2 3 4 5 6 7 8 9 10
Figure 3.6: A sequence of transpositions that sorts ω5 .
Let us define a sequence of transpositions τ1 , τ2 , . . . , τr+1 , as follows.
τi =





























τ (1, r + 2, r + 3)
i−2
τ (2 + i−2
3 , 2 + i, r + 4 + 2 3 )
i−3
τ (r − 2 3 , r + 2, 2r + 1 − i−3
3 )
i−4
i−4
τ (3 + 3 , r − 1 − 2 3 , 2r + 1 −
r+1
τ (1 + r+1
3 , 2 3 + 1, r + 2)
r+1
τ (1 + 3 , r + 2, 4 r+1
3 + 1)
r
2r
4r
τ ( 3 + 2, 3 + 2, 3 + 2)
τ ( 3r + 2, 2r
3 + 2, r + 2)
i=1
2 ≤ i ≤ r − 1, i ≡ 2 (mod 3)
2 ≤ i ≤ r − 1, i ≡ 0 (mod 3)
i−4
3 ) 2 ≤ i ≤ r − 1, i ≡ 1 (mod 3)
i = r, r ≡ 2 (mod 3)
i = r + 1, r ≡ 2 (mod 3)
i = r, r ≡ 0 (mod 3)
i = r + 1, r ≡ 0 (mod 3)
In Figure 3.6 it is demonstrated that this sequence of transpositions sorts ω 5 . We
prove by induction that this sequence of transpositions sorts ωr for all r. Let πi be
the permutation obtained after applying the first i transpositions to ωr , i.e., πi =
ωr · τ 1 · · · τ i .
Let us define δ to be equal to ωr (1). By the proof of Lemma 3.3.4 this definition is
equivalent to
´
³
(
r ≡ 2 (mod 3)
4 r+1
3
δ=
2r
r ≡ 0 (mod 3)
3 +1
Note that, for both cases 2δ ≡ 2r + 3 − δ (mod 2r + 1).
To help us describe πi at each step, we need the following definitions of subsequences
of ωr and ı. Let I(i, k) = [i i + 1 . . . i + k − 1]. Let I ∗ (i, k) = [i − k + 1 . . . i − 1 i]. Let
K(i, j, k) = [ωr (m) ωr (m+1) . . . ωr (m+k−1)], where ωr (m) = i, and ωr (m+k−1) = j.
Let B1 (k) = I(1, k), B2 (k) = I ∗ (2r + 2 − δ, k), B3 (k) = I(δ + 1, k), B4 (k) = I ∗ (δ, k),
B5 (k) = I(2r + 3 − δ, k), and B6 (k) = I ∗ (2r, k).
For 1 ≤ i ≤ r − 2 we shall show that πi is defined as follows:
If i ≡ 1 (mod 3) then πi = [B1 ((i − 1)/3 + 1) B4 ((i + 2)/3) B5 ((i + 2)/3) K((2 +
(i − 1)/3, δ − 1 − (i − 1)/3, r − 1 − i) B2 ((i + 2)/3) B3 ((i + 2)/3) K(2r + 3 − δ + (i +
2)/3, 2r + 2 − δ − (i + 2)/3, r − 2 − i) B6 ((i − 1)/3)].
Chapter 3. Sorting by Transpositions
61
If i ≡ 2 (mod 3) then πi = [B1 ((i + 1)/3 + 1) K(δ + 1 + (i + 1)/3, δ − (i + 1)/3, r −
1 − i) B2 ((i + 1)/3) B3 ((i + 1)/3) B4 ((i + 1)/3) B5 ((i + 1)/3 + 1) K(2 + (i + 1)/3, 2r +
2 − δ − (i + 1)/3, r − 2 − i) B6 ((i − 2)/3)].
If i ≡ 0 (mod 3) then πi = [B1 (i/3 + 1) K(δ + 1 + i/3, 2r + 1 − i/3, r − 1 − i) B4 (i/3 +
1) B5 (i/3 + 1) K((2 + i/3, δ − 1 − i/3, r − 2 − i) B2 (i/3 + 1) B3 (i/3) B6 (i/3 − 1)].
We shall prove this by induction. The base case is π1 . The first transposition is
τ1 = τ (1, r + 2, r + 3). This transposition moves 1 to the front of the permutation since
ωr (r + 2) = 1. Note also that π1 (2) = ωr (1) = δ, π1 (3) = ωr (2) = 2δ = 2r + 3 − δ,
π1 (4) = ωr (3) = 2, π1 (r + 1) = ωr (r) = 2r + 2 − 2δ = 2r + 1 − (2r + 3 − δ) = δ − 1,
π1 (r + 2) = ωr (r + 1) = 2r + 2 − δ, π1 (r + 3) = ωr (r + 3) = 1 + δ, π1 (r + 4) = ωr (r + 4) =
1 + 2δ = 2r + 4 − δ, and π1 (2r) = ωr (2r) = 2r + 1 − δ. So π1 = [1 δ 2r + 3 − δ K(2, δ −
1, r − 2) 2r + 2 − δ 1 + δ K(2r + 4 − δ, 2r + 1 − δ, r − 3)], which can be re-written as
[B1 (1) B4 (1) B5 (1) K(2, δ − 1, r − 2) B2 (1) B3 (1) K(2r + 4 − δ, 2r + 1 − δ, r − 3) B6 (0)],
which is the required form.
Now the induction step. Suppose πi is as described above. We shall apply transposition τi+1 and show that πi+1 is as described above. There are three different cases
that need to be considered.
Suppose i ≡ 1 (mod 3). Then τi+1 = τ (2 + (i − 1)/3, 2 + i, r + 4 + 2(i − 1)/3), so
applying this transposition results in the permutation [B1 ((i − 1)/3 + 1) K((2 + (i −
1)/3, δ − 1 − (i − 1)/3, r − 1 − i) B2 ((i + 2)/3) B3 ((i + 2)/3) B4 ((i + 2)/3) B5 ((i +
2)/3) K(2r + 3 − δ + (i + 2)/3, 2r + 2 − δ − (i + 2)/3, r − 2 − i) B 6 ((i − 1)/3)]. So we
can extend B1 and B5 by one. Note also that 2 + (i − 1)/3 + δ = δ + 1 + (i + 2)/3,
and 2r + 3 − δ + (i + 2)/3 + δ = 2r + 3 + (i + 2)/3. So we can rewrite the permutation
as [B1 ((i + 2)/3 + 1) K((δ + 1 + (i + 2)/3, δ − 1 − (i − 1)/3, r − 1 − (i + 1)) B 2 ((i +
2)/3) B3 ((i + 2)/3) B4 ((i + 2)/3) B5 ((i + 2)/3 + 1) K(2r + 3 + (i + 2)/3, 2r + 2 − δ −
(i + 2)/3, r − 2 − (i + 1)) B6 ((i − 1)/3)], which is the required form.
Suppose i ≡ 2 (mod 3). Then τi+1 = τ (r − 2(i − 2)/3, r + 2, 2r + 1 − (i − 2)/3), so
applying this transposition results in the permutation [B1 ((i + 1)/3 + 1) K(δ + 1 + (i +
1)/3, δ − (i + 1)/3, r − 1 − i) B4 ((i + 1)/3) B5 ((i + 1)/3 + 1) K(2 + (i + 1)/3, 2r + 2 −
δ − (i + 1)/3, r − 2 − i) B2 ((i + 1)/3) B3 ((i + 1)/3) B6 ((i − 2)/3)]. So we can extend B4
and B2 by one. Note also that δ > (i + 1)/3. So 2r + 1 − (i + 1)/3 + δ = δ − (i + 1)/3,
and δ − 1 − (i + 1)/3 + δ = 2r + 2 − δ − (i + 1)/3. So we can rewrite the permutation
as [B1 ((i + 1)/3 + 1) K(δ + 1 + (i + 1)/3, 2r + 1 − (i + 1)/3, r − 1 − (i + 1)) B 4 ((i +
1)/3 + 1) B5 ((i + 1)/3 + 1) K(2 + (i + 1)/3, δ − 1 − (i + 1)/3, r − 2 − (i + 1)) B 2 ((i +
1)/3 + 1) B3 ((i + 1)/3) B6 ((i − 2)/3)], which is the required form.
Suppose i ≡ 0 (mod 3). Then τi+1 = τ (3 + (i − 3)/3, r − 1 − 2(i − 3)/3, 2r + 1 − (i −
3)/3), so applying this transposition results in the permutation [B1 (i/3 + 1) B4 (i/3 +
1) B5 (i/3+1) K((2+i/3, δ −1−i/3, r −2−i) B2 (i/3+1) B3 (i/3) K(δ +1+i/3, 2r +1−
i/3, r − 1 − i) B6 (i/3 − 1)]. Clearly B3 and B6 can be extended by one. Note also that
δ +1+i/3+δ = 2r +4−δ +i/3 and 2r +1−δ −i/3+δ = 2r +1−i/3. So we can re-write
this permutation as [B1 (i/3 + 1) B4 (i/3 + 1) B5 (i/3 + 1) K((2 + i/3, δ − 1 − i/3, r − 1 −
Chapter 3. Sorting by Transpositions
62
(i+1)) B2 (i/3+1) B3 (i/3) K(2r +4−δ +i/3, 2r +1−δ −i/3, r −2−(i+1)) B 6 (i/3−1)],
which is the required form.
We now need to show that performing the last three transpositions results in the
identity permutation.
If r ≡ 2 (mod 3) then πr−1 = [B1 ((r + 1)/3) [δ + (r + 1)/3] B4 ((r + 1)/3) B5 ((r +
1)/3) B2 ((r + 1)/3) B3 ((r − 2)/3) B6 ((r − 2)/3 − 1)]. The next transposition is τr−1 =
τ (1 + (r + 1)/3, 2 + (r + 1)/3, 5(r + 1)/3 + 1). So πr−1 = [B1 ((r + 1)/3) B4 ((r +
1)/3) B5 ((r+1)/3) B2 ((r+1)/3) B3 ((r+1)/3) B6 ((r−2)/3−1)]. The next transposition
is τr = ((r + 1)/3 + 1, 2(r + 1)/3 + 1, r + 2), therefore πr = [B1 ((r + 1)/3) B5 ((r +
1)/3) B4 ((r+1)/3) B2 ((r+1)/3) B3 ((r+1)/3) B6 ((r−2)/3−1)]. The next transposition
is τr+1 = τ ((r + 1)/3 + 1, r + 2, 4(r + 1)/3 + 1). So πr+1 = [B1 ((r + 1)/3) B2 ((r +
1)/3) B5 ((r + 1)/3) B4 ((r + 1)/3) B3 ((r + 1)/3) B6 ((r − 2)/3 − 1)], which is the identity
permutation.
If r ≡ 0 (mod 3) then πr−2 = [B1 (r/3) B4 (r/3) B5 (r/3) [r/3 + 1] B2 (r/3)
B3 (r/3) B6 (r/3 − 1)]. The next transposition is τr−1 = τ (r/3 + 1, r + 1, 5r/3 +
2). So πr−1 = [B1 (r/3) [r/3 + 1] B2 (r/3) B3 (r/3) B4 (r/3) B5 (r/3) B6 (r/3 − 1)].
The next transposition is τr = τ (r/3 + 2, 2r/3 + 2, 4r/3 + 2). So πr = [B1 (r/3 +
1) B3 (r/3) B4 (r/3) B2 (r/3) B5 (r/3) B6 (r/3 − 1)]. The last transposition is τr+1 =
τ (r/3 + 2, 2r/3 + 2, r + 2). So πr+1 = [B1 (r/3 + 1) B4 (r/3) B3 (r/3) B2 (r/3) B5 (r/3)
B6 (r/3 − 1)] which is the identity permutation. 2
3.3.5
Strongly oriented cycles
Bafna and Pevzner define strongly oriented cycles to be oriented cycles that ‘have the
simplest “self-interleaving” structure among all oriented cycles’. They define a cycle
to be strongly oriented if it is oriented, and an interleaving transposition (that does
not act on any edges of the cycle) can transform the cycle into an unoriented cycle. A
cycle that is strongly oriented must be fully oriented. This is because it contains only
two grey edges directed to the right, and knots of length five (the only knots with only
two grey edges directed to the right) are not strongly oriented cycles. However, many
fully oriented cycles are not strongly oriented.
Bafna and Pevzner use strongly oriented cycles because they show that these cycles
admit 2-transpositions on their edges. However, we now have a better definition of such
cycles, the fully oriented cycles, that encompasses all cycles that admit 2-transpositions
on their edges, so we shall use fully oriented cycles instead of strongly oriented cycles.
Later in this chapter we prove several results on fully oriented cycles that are similar
to results Bafna and Pevzner proved about strongly oriented cycles.
3.3.6
Super oriented cycles
A cycle C is said to be super oriented if it is possible to apply a 2-transposition C
followed by a 2-transposition on one of cycles that C was split into.
Chapter 3. Sorting by Transpositions
63
If we could characterise super oriented cycles in the same way that we can characterise fully oriented cycles, then we may be able to improve the lower bounds for
transposition distance. Unfortunately, such a characterisation of super oriented cycles
has so far proved elusive. However, it is possible to detect some cycles that are not
super oriented, and for some permutations this helps us to improve the lower bound
for transposition distance. To describe this result we need an extra definition.
Define a grey edge (i, i + 1) to be crossing relative to the transposition τ (x, y, z) if
either:
(i) x ≤ π −1 (i) ≤ y − 1 and y ≤ π −1 (i + 1) ≤ z − 1, or
(ii) x ≤ π −1 (i + 1) ≤ y − 1 and y ≤ π −1 (i) ≤ z − 1.
This definition is motivated by the following lemma.
Lemma 3.3.6 Let τ be a transposition that transforms π into π 0 . Then a grey edge g
has different directions in tG(π) and tG(π 0 ) if and only if g is crossing relative to τ .
Proof We can use the transposition τ = τ (x, y, z) to partition the set {0, . . . , n + 1}
into three sets I1 , I2 , and I3 as follows. Let I1 = {π(i) : 0 ≤ i < x} ∪ {π(i) : z ≤ i ≤
n + 1}, I2 = {π(i) : x ≤ i < y}, and I3 = {π(i) : y ≤ i < z}. Now if i and i + 1 are in
the same set under this partitioning, then it is obvious that the direction of the grey
edge (i, i + 1) does not change as the result of applying τ . Similarly if i or i + 1 is in
I1 then it is easy to see that the direction of (i, i + 1) does not change as the result
of applying τ . In the remaining two cases (i, i + 1) is crossing relative to τ , and the
action of τ does change the direction of the grey edge. 2
Lemma 3.3.7 Let C be a cycle that contains only two grey edges that are directed
from left to right. Then C is not super oriented.
Proof Consider any oriented triple (x, y, z) of C, such that x < y < z. Then only one
grey edge of C is crossing relative to τ (x, y, z), and it must occur on the path between
z and x. By Lemma 3.3.6, this crossing edge is the only grey edge in C that changes
direction, and it becomes an edge directed to the right. The transposition τ (x, y, z)
splits cycle C into three cycles that each contain only one grey edge directed to the
right, so each cycle must be unoriented. Therefore C is not super oriented. 2
It is an open problem to determine if any cycle exists that is not super oriented,
but is fully oriented and has more than two grey edges directed to the right. Perhaps
such a cycle, if one exists, must produce a knot as the result of a 2-transposition on its
edges?
3.4
A new 3/2-approximation algorithm
The observations about even cycles, fully oriented cycles, and knots made in earlier
sections of this chapter, allow us to present a new 3/2-approximation algorithm that
Chapter 3. Sorting by Transpositions
64
is somewhat simpler than that presented by Bafna and Pevzner. We achieve the 3/2approximation bound by describing a sequence of transpositions that sorts a permutation and contains no (-2)-transposition, and at least two 2-transpositions for every
0-transposition.
Such a sequence of transpositions requires no more than 3/2 times the minimum
number of transpositions required to sort the permutation as predicted by Theorem
3.2.3. A useful definition for describing the 3/2-approximation algorithm is that of a
(0,2,2)-sequence. A (0, 2, 2)-sequence of transpositions is a sequence of three transpositions such that the first transposition is a 0-transposition, and the next two transpositions are 2-transpositions.
If for all permutations, apart from the identity permutation, we can find a 2transposition or a (0,2,2)-sequence, then we can achieve the approximation bound by
repeatedly using this result, until the permutation has been sorted. Now we already
know from the previous sections that we can apply a 2-transposition to a permutation if
its cycle graph contains a fully oriented cycle or an even length cycle. If tG(π) contains
a knot then we know that we can find a (0,2,2)-sequence, at least, on π.
So now we consider the case when tG(π) contains only odd length cycles that are
all unoriented. To study this case we need some extra definitions and some auxiliary
results.
Two pairs of black edges (bi , bj ) and (bk , bl ) are said to intersect if i < k < j < l,
or k < i < l < j. Similarly, a cycle intersects with two black edges x and y if the
cycle contains two black edges that intersect with x and y. Similarly, cycles C and D
intersect if C contains two black edges that intersect with two black edges of D.
Triples of black edges (bi , bj , bk ) and (bl , bm , bn ) interleave if i < l < j < m < k < n
or l < i < m < j < n < k. Similarly, cycles C and D interleave if C contains a triple
of black edges that interleave with a triple of black edges of D. Two transpositions
interleave if the triples of black edges that the transpositions act on interleave. A
transposition interleaves with a cycle C if the three edges that the transposition acts
on interleave with three black edges of C.
Bafna and Pevzner stated the following lemma for oriented cycles.
Lemma 3.4.1 Let (x, y, z) be a triple in a cycle C, and let u, v, w 6∈ C be black edges
of tG(π). Then τ (u, v, w) changes the orientation of triple (x, y, z) (i.e., it transforms
an oriented triple into an unoriented triple, and vice versa) if and only if (u, v, w)
interleaves with (x, y, z).
We prove several similar lemmas for fully oriented triples.
Lemma 3.4.2 Let (x, y, z) be a fully oriented triple in a cycle C, and let u, v, w 6∈ C
be black edges of tG(π). If (u, v, w) interleaves with (x, y, z) then τ (u, v, w) transforms
(x, y, z) into an unoriented triple in tG(π · τ ). Otherwise, (x, y, z) is a fully oriented
triple in tG(π · τ ).
65
Chapter 3. Sorting by Transpositions
1
...
a−1
a
...
b−1
b
...
c−1
c
...
l(C)
(i)
1
...
a−1
a
...
b−1
b
...
l(C)
(ii)
Figure 3.7: The oriented cycle obtained by an applying an transposition that interleaves
with an unoriented cycle.
Proof If the triples are interleaved then (x, y, z) is an unoriented triple in tG(π · τ )
by Lemma 3.4.1. Otherwise, after the transposition has been applied, x < y < z, or
y < z < x or z < x < y so (x, y, z) is an oriented triple by Proposition 3.3.2. Further,
(x, y, z) is a fully oriented triple, because the lengths of the paths connecting the edges
are no different than before. 2
Note that it is not true that an unoriented triple can be converted into a fully
oriented triple just by applying a transposition on edges that interleave with the unoriented triple. However, we can prove the following lemma about unoriented cycles.
Lemma 3.4.3 A transposition that interleaves with an unoriented cycle converts the
unoriented cycle into a fully oriented cycle.
Proof Let C be an unoriented cycle, and let τ be a transposition that interleaves with
C. Suppose that Cmin is to the left of the leftmost black edge that τ acts upon, and
Cmax is to the right of the rightmost black edge that τ acts upon. Then after applying
τ , C is a cycle with canonical labelling [1 l(C) . . . c b − 1 . . . a c − 1 . . . b a − 1 . . . ]
for some 1 < a < b < c < l(C) as shown in Figure 3.7 (i) (in which the black edges are
given their canonical labelling). Otherwise C becomes a cycle with canonical labelling
[1 b − 1 . . . a l(C) . . . b a − 1 . . . ] for some 1 < a < b < l(C) as is shown in Figure
3.7 (ii). In both cases the cycle is oriented, and in neither case is the cycle a knot, so
the cycle is fully oriented. 2
We now have enough auxiliary results to find a (0, 2, 2)-sequence when tG(π) contains only odd length unoriented cycles, and contains two interleaving cycles.
Lemma 3.4.4 Let π be a permutation, such that tG(π) contains only unoriented odd
cycles. Suppose that tG(π) contains two cycles, C and D, that interleave. Then there
exists a (0, 2, 2)-sequence of transpositions on π.
66
Chapter 3. Sorting by Transpositions
C
u
x
v
y
w
z
D
Figure 3.8: How C and D are interleaved initially.
Proof Let C be the cycle with the leftmost black edge. The two cycles are interleaved,
so let u, v and w be labels of edges of C and x, y and z be labels of edges of D such
that these triples interleave, and such that u is the edge bCmin . Further we may pick
x to be the leftmost edge of D to the right of u (so x is the edge bDmin ), y to be the
leftmost edge of D to the right of v, and z to be the leftmost edge of D to the right of
w. So tG(π) is as shown in Figure 3.8.
The transposition τ = τ (u, v, w) converts C into an unoriented cycle C 0 of tG(π · τ
and D into a fully oriented by Lemma 3.4.3.
We show that D 0 contains a fully oriented triple of edges that interleave with a
triple of edges of C 0 . If (y, x, z) is a fully oriented triple of D 0 then this is obvious. If
(y, x, z) is not a fully oriented triple of D 0 then, as a consequence of Lemma 3.3.2, two
of the paths connecting x, y and z must contain an odd number of black edges.
Suppose that it is the paths connecting x to z and y to x that both contain an odd
number of black edges. (Then since D 0 is odd the path connecting z to y contains an
even number of black edges.) Let s be the label of the first black edge on the path
connecting y to x. Then p(π, s) < p(π, v) since D is unoriented in tG(π), and y is the
leftmost edge of D to the right of v. So (y, s, z) is a fully oriented triple in tG(π · τ )
(Figure 3.9(a)).
Suppose that it is the paths connecting y to x and z to y that both contain an odd
number of black edges. Let s be the label of the first edge on the path connecting z to
y. Then p(π, s) < p(π, w) since D is unoriented in tG(π), and z is the leftmost edge of
D to the right of w. So (s, x, z) is a fully oriented triple in tG(π · τ ) (Figure 3.9(b)).
Suppose that it is the paths connecting z to y and x to z that both contain an odd
number of black edges. Let s be the label of the first edge on the path connecting x to
z. Then p(π, s) > p(π, z), since D is unoriented in tG(π), and x is the leftmost edge of
D. So (y, x, s) is a fully oriented triple in tG(π · τ ) as required (Figure 3.9(c)).
So D0 contains a fully oriented triple of edges that interleaves with C 0 . If we apply
the transposition on the fully oriented triple then C 0 will be a fully oriented cycle in
the cycle graph of the resulting permutation, by Lemma 3.4.3. So a 2-transposition can
67
Chapter 3. Sorting by Transpositions
C
y
x
z
s
D
(a)
C
y
s
z
x
D
(b)
C
y
x
z
s
D
(c)
Figure 3.9: How the cycles are interleaved after the 0-transposition.
Chapter 3. Sorting by Transpositions
68
then be performed on C 0 . Hence we can perform a (0, 2, 2)-sequence of transpositions
on π. 2
So now we only need to be able to find a (0,2,2)-sequence when all the cycles in
tG(π) are odd and unoriented, and no pair are interleaving. In fact we can find such a
(0,2,2)-sequence, but to prove this we require several auxiliary results. We begin with a
lemma proved by Bafna and Pevzner that shows that unoriented cycles must intersect
with other cycles.
Lemma 3.4.5 Let C be an unoriented cycle of tG(π). Let x and y be labels of two
black edges of C. Then there is a cycle D in tG(π) that intersects with x and y.
Lemma 3.4.6 Let C and D be unoriented cycles of tG(π), such that C and D intersect, and C has the leftmost black edge of the two cycles. Then the transposition
τ (Cmin , Dmin , Dmax ) converts C and D into a 1-cycle and an oriented cycle. Further,
the oriented cycle is fully oriented unless C and D are 3-cycles that interleave.
Proof Let x be the leftmost edge of C to the right of Dmin , and let y be the rightmost
edge of C to the left of Dmin . Then depending on whether Cmax < Dmax or Dmax <
Cmax the transposition τ (Cmin , Dmin , Dmax ) acts as shown in Figure 3.10 (i) or Figure
3.10 (ii). (Of course, the edge x could be the edge bCmax , and the edge y could be the
edge bCmin , and therefore the graphs would be not exactly as shown in the figure, but
in either case the resulting cycle decomposition is very similar to that shown.) The
transposition converts C and D into a 1-cycle and a long cycle C 0 . In both cases C 0 is
an oriented cycle, so C 0 is fully oriented unless it is a knot. If Dmax < Cmax then C 0
cannot be a knot because its canonical labelling begins [1 l . . . ], where l is the length
of the cycle. Otherwise, C 0 contains only two grey edges directed to the right (the edge
0
, and the edge between x and y). So C 0 could not be a knot of
that comes after bCmin
length greater than five, because such knots contain at least three grey edges directed
to the right.
Suppose that C 0 is a knot of length five. Then this cycle has canonical labelling
0
, and the edge with
[1 4 2 5 3]. The edge with canonical label 1 is the edge bCmin
canonical label 2 is the edge x because these edges precede the two grey edges that
are directed to the right. Let w be the edge with canonical label 4, y 0 be the edge
with canonical label 5 (this is the edge y if y differs from bCmin , and z be the edge
with canonical label 3. Now x < z < w in π · τ , and since the transposition doesn’t
change the relative positions of these edges it must be that x < z < w in π. So z must
be an edge of D because otherwise C would be an oriented cycle. So C and D are
interleaving 3-cycles. 2
Lemma 3.4.7 Let C and D be two unoriented cycles of tG(π), such that C and D
intersect, and C has the rightmost black edge of the two cycles. Then the transposition
τ (Dmin , Dmax , Cmax ) converts C and D into a 1-cycle and an oriented cycle. Further,
the oriented cycle is fully oriented unless C and D are 3-cycles that interleave.
69
Chapter 3. Sorting by Transpositions
C
y
x
D
C’
x
y
(i) Dmax > Cmax
C
y
x
D
C’
x
y
(ii) Dmax < Cmax
Figure 3.10: How the cycle graph changes.
70
Chapter 3. Sorting by Transpositions
C
D
c
C’
Figure 3.11: How tG(π) changes as the result of the transposition.
Proof The proof of Lemma 3.4.6 may be easily transformed to prove this lemma. 2
We now, as an aside, prove an interesting result about even length cycles.
Lemma 3.4.8 If tG(π) contains two unoriented even cycles C and D such that C and
D intersect, then it is possible to apply a sequence of two 2-transpositions on π.
Proof We can assume without loss of generality that C is the cycle that has the
leftmost black edge. The transposition τ (Cmin , Dmin , Dmax ) is a 2-transposition since
by Lemma 3.4.6 it transforms the two even cycles into two odd length cycles. By
Lemma 3.4.6 there is a fully oriented cycle in the resulting cycle decomposition, so we
can perform a further 2-transposition on this cycle. 2
Now back to auxiliary lemmas that will help us obtain the (0,2,2)-sequence that we
are seeking.
Lemma 3.4.9 Let C and D be two unoriented cycles of tG(π), such that C and D do
not intersect, and C has the leftmost black edge of the two cycles. Let c be the position
of a black edge of C. Then the transposition τ (c, Dmin , Dmax ) converts C and D into
a 1-cycle and an unoriented cycle.
Proof If Cmax < Dmin then the transposition τ (c, Dmin , Dmax ) acts as shown in Figure
3.11. (Note that, of course, c could be bCmin or bCmax , but these cases are similar to the
case illustrated.) So the transposition converts C and D into a 1-cycle and a proper
cycle C 0 . It is easy to verify that C 0 is an unoriented cycle, since it has only one grey
edge that is directed to the right. The case when Cmax > Dmax is similar. 2
Lemma 3.4.10 Let C and D be two unoriented cycles of tG(π), such that C and D
do not intersect, and D has the rightmost black edge of the two cycles. Let d be the
Chapter 3. Sorting by Transpositions
71
position of a black edge of D. Then the transposition τ (Cmin , Cmax , d) converts C and
D into a 1-cycle and an unoriented cycle.
Proof The proof of Lemma 3.4.9 may be easily transformed to prove this corollary.
2
Lemma 3.4.11 Let C and D be odd unoriented cycles of tG(π) that do not interleave,
and let C be the cycle with the leftmost black edge. Let τ (i, j, k) be a transposition that
acts on three of four black edges bCmin , bCmax , bDmin , bDmax , such that either:
(i) i = Cmin and j = Cmax , or
(ii) j = Dmin and k = Dmax .
Then τ (i, j, k) is a 0-transposition that converts C and D into a 1-cycle and a proper
cycle that is unoriented if C and D do not intersect, and is fully oriented otherwise.
Proof This lemma is a simple consequence of Lemmas 3.4.6, 3.4.7, 3.4.9 and 3.4.10.
2
We now have enough auxiliary results to obtain the (0,2,2)-sequence.
Lemma 3.4.12 Let π be a permutation such that tG(π) contains only unoriented odd
length cycles. Further suppose that no cycles in tG(π) interleave. Then there exists a
(0, 2, 2)-sequence of transpositions on π.
Proof Let C be the proper cycle in tG(π) with the leftmost black edge. By Lemma
3.4.5 another cycle must intersect with bCmin and bCmax . Let D be the cycle that
intersects with bCmin and bCmax , and that, among all the cycles that intersect with
these edges, contains the rightmost black edge to the left of bCmax . Let s be the second
leftmost black edge of C, and let t be the edge bCmax . (Note s 6= t, because l(C) ≥ 3).
If p(π, s) < Dmin then there must be a cycle in tG(π) that intersects with edges
bCmin and s of C (Lemma 3.4.5). Let E be a cycle that does intersect with these edges.
Now Emin > Cmin because C is the leftmost cycle of tG(π). Therefore, Cmin < Emin < s
and Emax > s because E intersects with bCmin and s. So either (a) s < Emax < Dmin ,
or (b) Dmin < Emax < t, or (c) t < Emax < Dmax , or (d) Dmax < Emax . These four
different ways that cycles D and E can be related, are shown in the top halves of
Figures 3.12 (a), (b), (c) and (d).
If p(π, s) > Dmin then there must be a cycle that intersects with edges s and bCmax
of C (by Lemma 3.4.5). Let E be a cycle that does intersect with these edges. Now
Emin > Cmin because C is the leftmost cycle of tG(π). Note that D cannot contain
black edges between s and t because otherwise it would interleave with C. Therefore
Emax < t because otherwise E would contradict the choice of D (since E would intersect
with C and contain a black edge to the left of bCmax that was further to the right than
any such edge of D). Therefore, Cmin < Emin < s and Emax > s because E intersects
72
Chapter 3. Sorting by Transpositions
C
C
E
D
E
D
t
s
s
t
C
C
t
s
D’
t
D’
(a)
s
(b)
C
E
C
E
D
D
t
s
t
s
C
C
t
t
s
D’
s
D’
(c)
(d)
D
D
C
E
E
C
s
t
s
t
C
C
t
s
t
s
D’
D’
(e)
(f)
Figure 3.12: The first transposition of a (0, 2, 2)-sequence.
with s and bCmax . So either (e) Dmin < Emin < s, or (f) Cmin < Emin < Dmin . These
two ways D and E can be related, are shown in at the top halves of Figures 3.12 (e)
and (f).
In Figure 3.12 a 0-transposition τ (by Lemma 3.4.11) is shown for each case. In
each case cycles D and E are converted into a 1-cycle and a long cycle D 0 . The
transposition shown interleaves with cycle C, so by Lemma 3.4.3 C is a fully oriented
cycle in the resulting cycle decomposition. In fact (bCmin , t, s) is a fully oriented triple
in the resulting cycle decomposition because all three paths connecting these edges are
even in length. We now explain why in each case it is possible to find a sequence of
two 2-transpositions on the resulting permutation.
(a) Cycles D and E do not intersect, so D 0 is unoriented by Lemma 3.4.11. Then
(bCmin , t, s) is a fully oriented triple of C that interleaves with D 0 so by Lemma 3.4.3 a
73
Chapter 3. Sorting by Transpositions
C
t
x
y
s
D’
Figure 3.13: Special case of (d).
sequence of two 2-transpositions can be performed on the permutation.
(b) Cycles D and E intersect, so D 0 is fully oriented by Lemma 3.4.11. Then
(bCmin , t, s) is a fully oriented triple of C that does not interleave with any fully oriented
0
0
triples of D 0 , since Cmin < Dmin
, and s > Dmax
. So by Lemma 3.4.2 a sequence of two
2-transposition can be performed.
(c) Cycles D and E intersect, so D 0 is fully oriented by Lemma 3.4.11. Then
(bCmin , t, s) is a fully oriented triple of C that does not interleave with any fully oriented
0
0
triples of D 0 , since Cmin < Dmin
, and s > Dmax
. So by Lemma 3.4.2 a sequence of two
2-transposition can be performed.
(d) If D and E do not intersect then D 0 is unoriented, and this case is similar to
(a). If D and E intersect then D 0 is fully oriented. Since D and E intersect, there must
be an edge x of E such that Dmin < x < Dmax . Further Cmax < x because otherwise
C and E would be interleaved. In particular, let x be the leftmost edge of E to the
right of bCmax . Let y be the edge that comes after x in D 0 . So tG(π · τ ) is as shown in
0
0
Figure 3.13. Then either (bDmin
, x, y) or (x, y, bDmax
) is a fully oriented triple that does
not interleave with (bCmin , t, s) which is a fully oriented triple of C. So by Lemma 3.4.2
a sequence of two 2-transposition can be performed.
(e) If D and E do not intersect then D 0 is unoriented, and this case is similar to
(a). If D and E intersect then D 0 is fully oriented. Since D and E intersect, there
must be an edge x of D such that Emin < x < Emax . Further x < s because otherwise
C and D would be interleaved. Let x be the rightmost edge of D to the left of s. Let
y be the edge that precedes x in D 0 . So tG(π · τ ) is as shown in Figure 3.14. Then
0
0
, y, x) or (y, x, bDmax
) is a fully oriented triple that does not interleave with
either (bDmin
(bCmin , t, s) which is a fully oriented triple of C. So by Lemma 3.4.2 a sequence of two
2-transposition can be performed.
(f) Cycles D and E intersect, so D 0 is fully oriented by Lemma 3.4.11. Then
(bCmin , t, s) is a fully oriented triple of C that does not interleave with any fully oriented
0
0
triples of D 0 , since Cmin < Dmin
, and t < Dmin
. So by Lemma 3.4.2 a sequence of two
2-transposition can be performed. 2
74
Chapter 3. Sorting by Transpositions
C
t
y
x
s
D’
Figure 3.14: Special case of (e).
We now state formally the argument that proves that all the lemmas in this section
allow us to describe a 3/2-approximation algorithm.
Lemma 3.4.13 For every permutation, except the identity permutation, we can find
either a 2-transposition or a (0, 2, 2)-sequence of transpositions on the permutation.
Proof By the definition of fully oriented cycles, a 2-transposition exists if tG(π)
contains a fully oriented cycle. By Lemma 3.2.5 a 2-transposition also exists if tG(π)
contains an even cycle. If tG(π) contains a knot then we can perform a (0, 2, 2)-sequence
of transpositions, by Lemma 3.3.5.
If tG(π) does not contain any fully oriented cycles, or even length cycles, or knots,
then tG(π) must contain only unoriented odd cycles. If tG(π) contains two interleaving
cycles then there is a (0, 2, 2)-sequence by Lemma 3.4.4. Otherwise tG(π) contains only
unoriented cycles, and no pair of these cycles are interleaving, so there is a (0, 2, 2)sequence by Lemma 3.4.12. 2
Theorem 3.4.1 There is a 3/2-approximation algorithm for sorting by transpositions.
Proof An entire sequence of transpositions that sorts a permutation can be generated
by re-applying Lemma 3.4.13 again and again until the permutation has been sorted.
This algorithm is illustrated in Figure 3.15. The sequence that is generated in this
way contains at least two 2-transpositions for every 0-transposition, and no (−2)transpositions. So the length of this sequence is at most 3/2 times the lower bound of
Theorem 3.2.3. Hence we have described a 3/2-approximation algorithm for sorting by
transpositions. 2
We now compare Bafna and Pevzner’s 3/2-approximation algorithm with the algorithm that has just been presented. In order to perform this comparison we briefly
introduce Bafna and Pevzner’s algorithm, which is summarised in Figure 3.16.
Chapter 3. Sorting by Transpositions
algorithm 3/2 Approximation(π: Permutation) is
τ , τ1 , τ2 , τ3 : Transposition;
begin
while π 6= ı loop
if tG(π) contains a fully oriented cycle C then
τ is a 2-transposition that acts on C;
π:= π · τ ;
elsif tG(π) contains an even cycle then
τ is a 2-transposition that acts on two even cycles (Lemma 3.2.5);
π:= π · τ ;
elsif tG(π) contains a knot C then
there is a (0, 2, 2)-sequence τ1 , τ2 , τ3 that acts on C (Lemma 3.3.5);
π:= π · τ1 · τ2 · τ3 ;
elsif tG(π) contains two interleaving cycles then
there is a (0, 2, 2)-sequence τ1 , τ2 , τ3 that acts on C (Lemma 3.4.4);
π:= π · τ1 · τ2 · τ3 ;
else
there is a (0, 2, 2)-sequence τ1 , τ2 , τ3 that acts on C (Lemma 3.4.12);
π:= π · τ1 · τ2 · τ3 ;
end if ;
end loop;
end 3/2 Approximation;
Figure 3.15: The new 3/2-approximation algorithm
75
Chapter 3. Sorting by Transpositions
76
algorithm BP Approximation(π: Permutation) is
begin
while π 6= ı loop
if tG(π) contains an oriented cycle C then
apply a 2-transposition or a (0,2,2)-sequence on π;
elsif tG(π) contains a long cycle then
if tG(π) contains two interleaving cycles then
apply a (0,2,2)-sequence on π (like Lemma 3.4.4);
else
apply a (0,2,2)-sequence on π (like Lemma 3.4.12);
end if ;
else
apply a sequence of two 2-transpositions;
end if ;
end loop;
end BP Approximation;
Figure 3.16: Bafna and Pevzner’s 3/2-approximation algorithm
There are two ways in which their algorithm may apply a (0,2,2)-sequence where
ours would apply a 2-transposition instead, so perhaps our algorithm will find shorter
sequences of transpositions. Their algorithm will apply either a 2-transposition or a
(0,2,2)-sequence on an oriented cycle, whereas our algorithm will always apply a 2transposition if one exists, and will apply a (0,2,2)-sequence only if the cycle is a knot.
Also, on occasion, their algorithm will apply (0,2,2)-sequences when the cycle graph
contains 2-cycles, whereas our algorithm would apply a 2-transposition on the even
cycles.
So we would expect our algorithm to sort permutations more efficiently. However
as will be shown in Section 3.6 there are some heuristics for sorting by transpositions
that we would expect to be even more efficient at solving instances of sorting by transpositions, even though the heuristics have no approximation guarantee. So, perhaps
more efficient algorithms exist anyway.
Unfortunately, although we can detect fully oriented cycles easily, we know of no
quicker means of finding a fully oriented triple in a fully oriented cycle than trying
every triple of edges in the cycle. So finding a strongly oriented triple has O(n3 )
worst case time-complexity. The other steps inside the main loop of our algorithm
have the following worst case time-complexities: building tG(π) is O(n), determining if
tG(π) contains a fully oriented cycle is O(n), determining if tG(π) contains interleaving
cycles is O(n2 ), and performing a transposition is O(n). Hence, the overall worst case
time-complexity of the algorithm is O(n4 ), because we need to perform at most O(n)
Chapter 3. Sorting by Transpositions
77
transpositions. However we believe that it may be possible to improve the worstcase complexity of this algorithm, by finding a more clever algorithm for finding fully
oriented triples in fully oriented cycles, or by improving the algorithmic analysis above.
Bafna and Pevzner claim that their algorithm is O(n2 ), but it seems to be O(n3 )
since before each transposition it must determine if any two cycles are interleaving,
which is O(n2 ), and it must perform O(n) transpositions. They are able to find transpositions on oriented cycles in O(n) because they accept a (0, 2, 2)-sequence even when
a 2-transposition exists.
We prove some results that are similar to results proved by Bafna and Pevzner.
Our proof of Lemma 3.4.12 is simpler than their proof of a similar result, because it
involves fewer cases, and does not involve arguments based on symmetries. To be fair
though, their proof of a result like Lemma 3.4.4 based on strongly oriented cycles is
shorter and perhaps simpler than our proof, though ours is more precise.
It is worth noting that Guyer, Heath and Vergara [GHV95] were “unable to” implement Bafna and Pevzner’s 3/2-approximation algorithm because they could not resolve
some “implementation details.” Therefore a new 3/2-approximation algorithm that is
simpler and is proved in a more precise manner is a useful contribution.
3.5
3.5.1
A tighter lower bound
Hurdles
In this section improved lower bounds for transposition distance are presented. These
improved lower bounds are achieved by identifying subgraphs of the cycle graph that
are difficult to sort. These difficult subgraphs are called hurdles, by analogy with the
terminology of Hannenhalli and Pevzner for sorting by reversals. Below we define
hurdles and then we describe some improved lower bounds for transposition distance
based on counting hurdles.
Let us construct an overlap graph tH(π) that contains a vertex for each cycle in
tG(π). Connect two vertices by an edge if their respective cycles intersect. Collections
of cycles of tG(π) that form components in tH(π) will be of interest. We shall call a
collection of cycles of tG(π) a component of π if the collection consists of all the cycles
in one component of tH(π).
Example Let π be the permutation [8 10 9 11 1 6 5 4 3 2 7]. Then tG(π) contains
four cycles: C1 = (12, 1, 5), C2 = (11, 9, 7), C3 = (10, 8, 6), and C4 = (4, 2, 3), as shown
in Figure 3.17. The components of π are {C1 }, {C2 , C3 }, and {C4 }. 2
Let C be a component of tG(π). Now suppose that we obtain a graph from tG(π)
by deleting all edges and vertices that are not part of C, and then merging vertices that
are next to each other in the resulting graph, as we draw it, but are not adjacent by a
black edge. For example, if C is the component {C1 } in Figure 3.17, we would merge
vertices 8 and 11, and 7 and 1. This resulting graph can have its vertices relabelled
78
Chapter 3. Sorting by Transpositions
0
8
10
9
11
1
6
5
4
3
2
7
12
Figure 3.17: A cycle graph with three components.
so that the leftmost vertex is labelled 0, and each grey edge is directed from vertex i
to vertex i + 1. Reading vertices from left to right we obtain a permutation called the
component permutation πC .
If C contains no even cycles and πC cannot be sorted using only 2-transpositions
then we say that C is a hurdle. If πC can be sorted using only 2-transpositions then C
is a sortable component (whether C contains an even cycle or not). Let th(π) denote
the number of hurdles in tG(π). A transposition τ clears a hurdle C if no part of any
cycle in C is part of a hurdle in tG(π · τ ).
Example For the example permutation above, π{C1 } = [2 1], π{C2 ,C3 } = [5 4 3 2 1]
and π{C4 } = [2 1]. Components {C1 } and {C4 } are sortable components, but the
component {C2 , C3 } is a hurdle. 2
It is possible to improve the lower bound of Theorem 3.2.3 by counting hurdles in
tG(π). Bafna and Pevzner’s lower bound is based on assuming that each transposition
in a minimal length sorting sequence is a 2-transposition. Note that a 2-transposition
on a cycle in one component does not affect cycles in other components. Also, if a
2-transposition acts on two cycles then both cycles must be even in length. So if tG(π)
contains a hurdle then at least one transposition that is not a 2-transposition will be
required to sort π. In fact we can improve the lower bound as shown below.
Theorem 3.5.1 For every permutation π, td(π) ≥ ((n + 1 − tcodd (π))/2 + dth(π)/2e.
Proof All hurdles of tG(π) must be cleared by at least one 0-transposition, or (−2)transposition. A transposition can act on at most three hurdles, so a single transposition can clear at most three hurdles. If a transposition clears three hurdles then
it must merge three cycles, from different components, together into one cycle. Such
a transposition is a (−2)-transposition. So for every transposition that clears three
hurdles, an extra 2-transposition is required. In fact hurdles can be cleared more
efficiently by clearing two at a time. A 0-transposition on two cycles in different components may clear two hurdles. For example, the first transposition in Figure 3.18 is a
0-transposition that clears two hurdles. Clearing hurdles in pairs with 0-transpositions
in this way requires dth(π)/2e 0-transpositions to clear all the hurdles. 2
A problem with this bound based on hurdles is that detecting hurdles may not
be any easier than determining the transposition distance of a given permutation.
79
Chapter 3. Sorting by Transpositions
However, there are some hurdles that can be detected easily. These hurdles consist
only of odd length cycles that are not fully oriented. Let thd (π) represent the number
of detectable hurdles like this in π. The lower bound that we obtain using this count
of hurdles is as follows:
Theorem 3.5.2 For every permutation π, td(π) ≥ (n + 1 − tcodd (π))/2 + dthd (π)/2e.
Proof Clearly thd (π) ≤ th(π). The theorem then follows from Theorem 3.5.1. 2
Let tl(π) = (n + 1 − tcodd (π))/2 + dthd (π)/2e. In Section 1.1, we observed that
td(π) = td(π −1 ). Let π = [7 5 3 1 8 6 4 2], so that π −1 = [4 8 3 7 2 6 1 5]. Now
tl(π) = 3, but tl(π −1 ) = 4, and given the above identity it must be that td(π) ≥ 4.
This example inspires the following lower bound:
Theorem 3.5.3 For all permutations π, td(π) ≥ max{tl(π), tl(π −1 )}.
Proof This theorem can be derived easily from Theorem 3.5.2 and the identity td(π) =
td(π −1 ). 2
We have found that this lower bound is useful to speed up an implemented branch
and bound algorithm for sorting by transpositions. The bound can also be used to
obtain permutations that have transposition distances arbitrarily far away from the
lower bound of Theorem 3.2.3, as we show in the next section.
3.5.2
How tight are the lower bounds?
We now show that Bafna and Pevzner’s lower bound for transposition distance (Theorem 3.2.3) can be arbitrarily far away from the exact transposition distance. We need
some extra definitions to describe permutations with this property.
If π is a permutation of the set {1, . . . , n} and k is an integer then we define π + k
to be the permutation of the set {k + 1, . . . , k + n} obtained by adding k to each of the
elements of π. In particular this means that (π + k)(i) = π(i) + k, for 1 ≤ i ≤ n.
If π is a permutation of the set {1, . . . , n} and φ is a permutation of the set
{n + 1, . . . , m} then we can define π ++ φ to be the permutation of the set {1, . . . , m}
obtained by appending φ to π. For example if π = [2 4 3 1] and φ = [6 5] then
π ++ φ = [2 4 3 1 6 5].
We now build a permutation κk that has transposition distance arbitrarily distant from the lower bound of Theorem 3.2.3. Let κ = [4 3 2 1]. Define κ 1 = κ,
and κk = κk−1 ++ [5(k − 1)] ++ (κ + 5(k − 1)), for k ≥ 2. So, for example, κ3 =
[4 3 2 1 5 9 8 7 6 10 14 13 12 11]. We prove that κk has the property that we want.
Theorem 3.5.4 The transposition distance of κk is given by the formula
td(κk ) = 2k +
k
n + 1 − tcodd (κk )
k
=
.
+
2
2
2
» ¼
» ¼
80
Chapter 3. Sorting by Transpositions
0
4
3
2
1
5
9
8
7
6
10
0
1
5
4
3
2
9
8
7
6
10
0
1
2
9
5
4
3
8
7
6
10
0
1
2
3
8
9
5
4
7
6
10
0
1
2
3
4
7
8
9
5
6
10
0
1
2
3
4
5
6
7
8
9
10
Figure 3.18: An optimal length sorting of κ2 .
Proof The cycle graph of κk consists of k knots of size five that do not intersect with
each other. Therefore, by Theorem 3.5.1, the transposition distance of κ k is at least
(n + 1 − tcodd (κk ))/2 + dk/2e.
It is possible to sort κk by sorting pairs of knots together as shown in Figure 3.18. If
k is odd, then one knot must also be sorted on its own. The sequence of transpositions
that we have just described has the length given by the above formula. 2
In fact we can generalise the above theorem as follows. Let π be a permutation
that contains only odd length cycles and for which td(π) ≥ (n + 1 − tcodd (π))/2 + 1.
Let π1 = π, and πk = πk−1 ++ [(k − 1)(n + 1)] ++ (π + (k − 1)(n + 1)). Then td(πk ) ≥
(n + 1 − tc(πk ))/2 + dk/2e.
So we have shown that κk is an example of a family of permutations whose transpo-
Chapter 3. Sorting by Transpositions
81
sition distance can be arbitrarily far from the lower bound of Theorem 3.2.3. However
it appears that such permutations are rare, and the bound of Theorem 3.2.3 is a good
bound in practice (Section 3.6).
Note also that td(κ1 ) = 3, but the lower bound for κ1 is only 2. So any approximation algorithm that has its approximation ratio based only on the lower bound of
Theorem 3.2.3 cannot have an approximation ratio better than 3/2.
Of course, the lower bound of Theorem 3.5.1 based on hurdles is exact for κ k . So,
it is natural to ask if are there any permutations that can be arbitrarily distant from
this lower bound? In fact this question remains open, however we conjecture that the
answer is yes for reasons explained below.
Consider the permutation π1 = [3 2 1 6 5 4 9 8 7 12 11 10]. The lower bound based
on hurdles (Theorem 3.5.1) is 5 transpositions, but td(π1 ) = 6. (The exact distance
of this permutation was calculated using a branch and bound algorithm described in
section 3.6.) Note that the overlap graph for this permutation tH(π) contains only
one component, but that component is not a hurdle or a sortable component. This
permutation demonstrates a way in which the hurdles lower bound can underestimate
the transposition distance, because of the way it deals with components containing
even cycles.
Consider also the permutation π2 = [5 4 3 8 7 6 11 10 9 14 13 12 17 16 15 2 1]. The
cycle graph, tG(π2 ) contains six 3-cycles that are all unoriented, such that no pair of the
cycles are interleaving. The overlap graph, tH(π2 ) contains only one component, that
is a hurdle. In can be shown that td(π2 ) = 9 (using the branch and bound algorithm)
but the lower bound of Theorem 3.5.1 is 7. So π2 is a permutation distance 2 greater
than the lower bound based on hurdles.
Note also, that the permutation π2 demonstrates a manner in which hurdles of the
transposition cycle graph are unlike hurdles of the reversal cycle graph, because the
hurdle in the transposition graph requires more than one transposition in order to be
cleared, whereas hurdles of the reversal cycle graph require only one reversal to be
cleared (with some special permutations, fortresses, requiring an extra reversal).
We conjecture that it may be possible to build permutations that have transposition
distances arbitrarily far from the lower bound of Theorem 3.5.1 by concatenating permutations like π1 and π2 together, or building larger permutations that have structures
similar to those of π1 and π2 .
3.5.3
The transposition diameter
The transposition diameter tD(n) is the maximum value of td(π) taken over all permutations π of length n. Bafna and Pevzner proved that tD(n) ≤ 3n/4, since their
3/2-approximation algorithm sorts any permutation using at most 3/4.(n + 1 − c odd (π))
transpositions. They also discovered that, for 3 ≤ n ≤ 10, tD(n) = bn/2c + 1, and that
this value is achieved by the reverse permutation, Rn . They also stated that, for all n,
td(Rn ) ≤ bn/2c + 1. We prove that td(Rn ) = bn/2c + 1, and make a conjecture about
Chapter 3. Sorting by Transpositions
82
transposition diameter.
For odd n ≥ 3 we define the broken reverse permutation, BRn , to be the permutation
that results from applying the transposition τ (1, (n+1)/2, (n+1)/2+2) to R n . If n = 3
then BRn = [2 1 3], and for larger values of n then BRn = [dn/2e bn/2c n . . . dn/2e +
1 bn/2c − 1 . . . 1].
Lemma 3.5.1 For odd n, td(BRn ) ≤ bn/2c.
Proof We prove this lemma by induction. The base case is when n = 3. Clearly one
transposition is enough to sort [2 1 3].
Now suppose that the lemma is true for all n ≤ 2k + 1. Let n = 2k + 3. Now apply
the transposition τ (2, (n + 1)/2 + 1, (n + 1)/2 + 3). If n = 5, we obtain the permutation
π = [3 4 1 2 5], and otherwise we obtain the permutation π = [dn/2e dn/2e + 1 bn/2c −
1 bn/2c n . . . dn/2e + 2 bn/2c − 2 . . . 1]. Now gl(π) = gl(BR2k+1 ), so by the inductive
hypothesis, and Theorem 3.2.2, td(π) ≤ k. Therefore td(BR2k+3 ) ≤ k + 1, and we have
proved the lemma by induction. 2
We now prove the following theorem, which is stated without proof by Bafna and
Pevzner.
Theorem 3.5.5 td(Rn ) ≤ bn/2c + 1, for all n ≥ 2.
Proof If n is even, we make the transposition τ (1, 2, n + 1) to move the element n to
its correct position. We then treat what remains as Rn−1 . The fact that, for n even,
bn/2c + 1 = 1 + (b(n − 1)/2c + 1) means that the truth of the result for n even will
follow if we establish its validity for n odd.
For n odd, perform the transposition τ (1, (n + 1)/2, (n + 1)/2 + 2). The resulting permutation is BRn . By Lemma 3.5.1 we may sort this permutation with bn/2c
transposition. So Rn can be sorted using bn/2c + 1 transpositions. 2
In fact, for n odd we can prescribe a sequence of transpositions, τ1 , . . . , τr (where
r = (n + 1)/2), that sorts Rn as follows:
τi =
(
τ (i, i + (n − 1)/2, i + (n + 3)/2), i = 1, . . . , r − 1
τ (1, (n + 1)/2, n),
i = r.
In Figure 3.19 this sequence of transpositions is applied to R11 .
Theorem 3.5.6 td(Rn ) = bn/2c + 1, for all n ≥ 2.
Proof By Theorem 3.5.5 td(Rn ) ≤ bn/2c + 1.
If n ≡ 3(mod 4) then tG(Rn ) consists of only 2 even length cycles. So by Theorem
3.2.3 it is impossible to sort Rn using fewer than bn/2c + 1 transpositions.
If n ≡ 1(mod 4) then tG(Rn ) consists of only 2 odd length cycles. Both cycles are
unoriented, so again it is impossible to sort Rn using fewer than bn/2c+1 transpositions.
Chapter 3. Sorting by Transpositions
83
11 10 9 8 7 6 5 4 3 2 1
6 5 11 10 9 8 7 4 3 2 1
6 7 4 5 11 10 9 8 3 2 1
6 7 8 3 4 5 11 10 9 2 1
6 7 8 9 2 3 4 5 11 10 1
6 7 8 9 10 1 2 3 4 5 11
1 2 3 4 5 6 7 8 9 10 11
Figure 3.19: A sequence of transpositions that sorts R11 .
However, when n is even, tG(Rn ) consists of one oriented odd length cycle. So in
this case the lower bound of Theorem 3.2.3 is only bn/2c transpositions. However, the
cycle in tG(Rn ) has canonical labelling [1 n n−2 . . . 2 n+1 n−1 . . . 3]. Therefore, by
Lemma 3.3.7, the cycle is not super oriented since it contains only two grey edges that
are directed to the right. So it is impossible to find a sequence of two 2-transpositions on
Rn , and therefore it is impossible to sort Rn using fewer than bn/2c + 1 transpositions.
2
As a result of Theorem 3.5.6, bn/2c + 1 ≤ tD(n) ≤ 3n/4. In fact, we conjecture
that the reverse permutation achieves the transposition diameter for all n.
Conjecture 3.5.1 For all n, tD(n) = bn/2c + 1.
We make this conjecture because, on the one hand, the more cycles there are in
tG(π) the fewer 2-transpositions are needed to sort π by the lower bound of Theorem
3.2.3, and on the other hand, the fewer cycles there are in tG(π), then the longer these
cycles must be and so the more likely it is that the cycles are fully oriented, or are
knots, and hence the more likely it is that a sequence of many 2-transpositions can
be applied to π. We believe these two factors together make it impossible to have
td(π) > bn/2c + 1. Some evidence supporting this conjecture is presented in Section
3.6.
3.6
Solving instances of sorting by transpositions in practice
In order to investigate how well instances of sorting by transpositions can be solved in
practice, several approximation algorithms and an exact algorithm were implemented
and tested. The various algorithms use heuristics based on results of the previous
sections. We describe the algorithms in Section 3.6.1. The algorithms were tested with
Chapter 3. Sorting by Transpositions
84
various kinds of permutation that are described in Section 3.6.2. Tables of raw data
describing each algorithm’s performance are presented in Section 3.6.3. In Section 3.6.4
we analyse these tables, and try to explain the observed behaviour. We conclude that
most instances of sorting by transpositions can be solved optimally, at least for the
range of sizes covered by our experiments.
3.6.1
The algorithms
The different algorithms that were implemented are described below. Each description
begins with the name of the algorithm, followed by an explanation of the heuristics the
algorithm uses and a statement about the complexity of the algorithm.
lower bound
This algorithm is not an approximation algorithm or exact algorithm for sorting by
transpositions. Instead, it simply calculates the lower bound for transposition distance
based on detectable hurdles and the inverse permutation (Theorem 3.5.3). To calculate
this bound the algorithm must build the cycle graph tG(π) and the overlap graph
tH(π). Counting the number of cycles in tG(π) can be performed in O(n) steps, but
building tH(π), and identifying its components takes O(n2 ) steps. So this algorithm
is O(n2 ).
approx
This algorithm uses several heuristics, based on the results of the previous sections, to
find a sequence of transpositions that sorts a given permutation. The algorithm has
no approximation guarantee, but in practice we expect it to perform better than the
3/2-approximation algorithm described in Section 3.4, for reasons explained below.
The algorithm is described in Figure 3.20. The first step of the algorithm checks if
the permutation is transposition equivalent to the reverse permutation and sorts the
permutation optimally if it is. Otherwise a sequence of steps is followed that is similar
to the sequence of steps used in the 3/2-approximation algorithm.
If tG(π) contains a strongly oriented cycle then a 2-transposition τ is applied on a
shortest strongly oriented cycle C, splitting C into three cycles C1 , C2 and C3 . Note
that, in particular, τ is chosen so that the size of C1 is maximised, and then so that
the size of C2 is maximised. This strategy was implemented because, by Lemma 3.3.3,
larger cycles are more likely to be strongly oriented than smaller cycles. Therefore
when we have a choice between applying a 2-transposition on a small cycle or a 2transposition on a large cycle, we apply the transposition on the small cycle, because
it is less likely that the small cycle should have been fully oriented, and it is more
likely that the large cycle will remain fully oriented in the cycle graph of the resulting
permutation. We choose the transposition that maximises C1 and then C2 because we
are attempting to maximise the probability that these cycles are fully oriented.
Chapter 3. Sorting by Transpositions
85
algorithm approx(π: Permutation) is
τ : Transposition;
ts: array of Transposition;
i: Integer;
begin
i:= 0;
while π 6= ı loop
i:= i+1;
if gl(π) = R|gl(π)| then
sort π optimally from here (Theorem 3.5.6);
exit loop;
else
if tG(π) contains a fully oriented cycle then
τ := a 2-transposition that acts on a smallest fully oriented cycle;
elsif tG(π) contains interleaving even cycles then
τ := a 2-transposition that acts on two interleaving even cycles (Lemma 3.4.8);
elsif tG(π) contains even cycles then
τ := a 2-transposition that acts on two even cycles (Lemma 3.2.5);
elsif tG(π) contains a knot then
τ := the 0-transposition from the (0, 2, 2)-sequence of Lemma 3.3.5;
elsif tG(π) contains two interleaving cycles then
τ := the 0-transposition from the (0, 2, 2)-sequence of Lemma 3.4.4;
else
τ := the 0-transposition from the (0, 2, 2)-sequence of Lemma 3.4.12;
end if ;
π:= π · τ ;
ts(i):= τ ;
end if ;
end loop;
return i, ts;
end approx;
Figure 3.20: The algorithm: approx
Chapter 3. Sorting by Transpositions
86
The algorithm does not achieve the 3/2-approximation bound because we have no
guarantee that it applies two 2-transpositions after every 0-transposition. But the
strategy for selecting 2-transpositions is designed with the hope that it will find a long
sequence of 2-transpositions, if one exists, and so we hope that the algorithm will find
at least two 2-transpositions after every 0-transposition. The algorithm performs so
well in practice that implementing the 3/2-approximation algorithm did not seem to
be worthwhile.
The algorithm has O(n4 ) worst-case time complexity by the same reasons that the
3/2-approximation algorithm has this complexity. In practice, though, the runtime of
the algorithm does not appear to increase as rapidly as this.
alt approx
This algorithm uses the same strategy as approx except that when a fully oriented
cycle exists, the algorithm favours 2-transpositions on odd length cycles before 2transpositions on even length cycles.
The reason for this variation is that we know, by Lemma 3.2.5, that if tG(π) contains
even cycles then we can perform a 2-transposition on π, whether the even cycles are
fully oriented or not. So the probability of being able to apply a 2-transposition on an
odd cycle could be thought of as being much smaller than the probability of applying
a 2-transposition on even cycles.
Another way of thinking about this algorithm is that it keeps 2-transpositions on
even cycles in reserve until it gets stuck trying to apply 2-transpositions on odd cycles.
The time complexity of this algorithm is O(n4 ) just like approx.
tlis
This algorithm is based on an algorithm described by Guyer, Heath and Vergara
[GHV95]. The algorithm repeatedly applies a transposition τ to π such that π · τ
has the longest possible increasing subsequence. This algorithm is described in Figure
3.21. A modification we have made to this algorithm (that is not indicated in the
figure) is that the transposition is chosen so that it maximises the length of the smaller
block moved, over all transpositions that increase the longest increasing sequence maximally. Without this modification, if the length of the longest increasing subsequence
can be increased by only one element then the algorithm could repeatedly move only
a single element in the permutation until the permutation is sorted, which is likely to
be a very inefficient method of sorting the permutation.
There are O(n3 ) transpositions that need to be considered on each iteration of the
loop, and for each transposition the length of the longest increasing subsequence is
calculated in O(n log n) time. The loop is iterated O(n) times, so the overall timecomplexity of the algorithm is O(n5 log n).
Chapter 3. Sorting by Transpositions
algorithm tlis(π: Permutation) is
algorithm lis(π: Permutation) is
begin
return the length of a longest increasing subsequence of π;
end lis;
τ : Transposition;
ts: array of Transposition;
i: Integer;
begin
i:= 0;
while π 6= ı loop
τ := a transposition such that lis(π · τ ) is maximised;
π:= π · τ ;
i:= i+1;
ts(i):= τ ;
end loop;
return i, ts;
end tlis;
Figure 3.21: The algorithm: tlis
87
Chapter 3. Sorting by Transpositions
88
algorithm random(π: Permutation) is
τ : Transposition;
ts: array of Transposition;
i: Integer;
begin
i:= 0;
while π 6= ı loop
i:= i+1;
if π admits a 2-transposition then
τ := a 2-transposition chosen uniformly at random;
elsif tG(π) contains a knot then
τ := the 0-transposition from the (0, 2, 2)-sequence of Lemma 3.3.5;
elsif tG(π) contains two interleaving cycles then
τ := the 0-transposition from the (0, 2, 2)-sequence of Lemma 3.4.4;
else
τ := the 0-transposition from the (0, 2, 2)-sequence of Lemma 3.4.12;
end if ;
π:= π · τ ;
ts(i):= τ ;
end loop;
return i, ts;
end random;
Figure 3.22: The algorithm: random
random
This algorithm repeatedly selects a 2-transposition uniformly at random from among
all the 2-transpositions that are possible on the permutation until the permutation
is sorted. If no 2-transposition is possible then a 0-transposition is applied using the
same strategy that approx uses. The algorithm is summarised in Figure 3.22. This
algorithm has O(n4 ) complexity because in the worst case counting the total number
of 2-transpositions on π is an O(n3 ) operation, and we may need to perform O(n)
transpositions. Note that we have optimised the way we count 2-transpositions so
that, in practice, this algorithm appears to have a better runtime complexity.
rpt random
This algorithm repeatedly applies the algorithm random until an exact solution is found,
or a specified amount of processing time has elapsed, in which case the algorithm returns
the shortest sequence of transpositions that was obtained within the time-limit. The
algorithm uses alt approx to produce the initial sorting sequence. Of course, an
Chapter 3. Sorting by Transpositions
89
application of random is abandoned as soon as it cannot improve upon the best sorting
sequence found so far.
branch
This algorithm uses branch-and-bound search to find the exact transposition distance
of a given permutation. This algorithm is summarised in Figure 3.23. The lower
bound used by the algorithm is the bound based on detectable hurdles and the inverse
permutation (Theorem 3.5.3). The search is bounded by a specified time limit. Of
course, there are permutations for which this algorithm will not be able to find the exact
solution in the specified time. If time runs out during the search then the algorithm
returns the shortest sequence of transpositions that was obtained during the time-limit.
best
This algorithm tries all the algorithms described above (except for lower bound) and
selects the shortest sorting sequence found. Note that this algorithm also tries a few
variants of the algorithm branch that can sometimes establish the exact transposition
distance of a permutation faster than branch. However in the experiments these variant algorithms could only determine the exact transposition distance of a handful of
permutations that the other algorithms could not establish. Furthermore, these variant
algorithms could not establish the transposition distance of many permutations that
the other algorithms could. Therefore we do not include tables for these algorithms, or
describe these algorithms fully, though we do describe the variants a little in Section
3.6.4.
3.6.2
Permutations
The algorithms were tested with three different kinds of permutations. Of course, we
test the algorithms with random permutations (random permutations). We also test
the algorithms with two kinds of permutations that have cycle graphs containing only
3-cycles. We are interested in these permutations because they are important when
we consider the concept of transposition tightness which is defined by analogy with
reversal tightness (Section 2.5). We say that a permutation π is transposition tight if
it is possible to sort π by applying a sequence of transpositions to π such that each
transposition removes three breakpoints. If a permutation is transposition tight then it
must have a cycle graph containing only 3-cycles. We generate permutations containing
only 3-cycles in their cycle graph that may or may not be transposition tight (3-cycle
permutations), as well as permutations that are transposition tight (transposition tight
permutations).
We describe the three kinds permutations in more detail below.
Chapter 3. Sorting by Transpositions
algorithm branch(π: Permutation) is
bound: Integer
current, best: array of Transposition;
algorithm search(π, depth) is
begin
if π = ı then
if depth < bound then
bound:= depth;
best:= current;
end if ;
else
for each transposition τ in order of decreasing ∆tcodd (π, τ ) loop
if lower bound(π · τ ) + depth + 1 < bound then
current(depth + 1):= τ ;
search(π · τ, depth + 1);
end if ;
end loop;
end if ;
end search;
begin
bound, best:= alt approx(π);
search(π, 0);
return bound, best;
end branch;
Figure 3.23: The algorithm: branch
90
Chapter 3. Sorting by Transpositions
91
random permutations
These are generated uniformly at random from among all strip free permutations of a
given length.
3-cycle permutations
These are generated uniformly at random from among all permutations of a given
length that have cycle graphs containing only 3-cycles such that no 3-cycle contains
both bi and bi+1 for any i. Note that these permutations can only be generated if n ≡ 2
(mod 3).
We generate these permutations by randomly partitioning {b1 , . . . , bn+1 } into sets
of size three, such that no set contains bi and bi+1 for any i. Each of these sets represents
a potential 3-cycle in the cycle graph. We then randomly decide if each potential 3cycle is oriented or unoriented. Having made these decisions we can add grey edges
to the black edges, so as to obtain a candidate cycle graph. We then attempt to label
all the vertices in this graph by starting from vertex 0 and following grey edges in the
graph (since each grey edges is directed from vertex i to vertex i + 1). (Of course, we
can identify vertex 0 to begin with because it is adjacent to b1 .) If we are able to label
all the vertices in the graph, then a permutation π exists with this candidate cycle
graph and it is easy to read π from the graph. If we cannot label all the vertices then
we redo the entire process until it works.
Note that if a 3-cycle contains black edges bi and bi+1 , for some i, then that 3-cycle
must be fully oriented. Such a 3-cycle remains fully oriented until a 2-transposition is
applied on the cycle, or a 0-transposition or (-2)-transposition is applied that acts on at
least one of the edges of the cycle. Furthermore, if we apply a 2-transposition on such
a cycle, the overlapping pattern of the other cycles in the cycle graph is unaffected.
We conjecture that if tG(π) contains such a cycle, we may always obtain an optimal
length sequence of transpositions by initially applying a 2-transposition on this cycle.
In fact, this conjecture holds when the permutation is transposition tight. Therefore
we believe that these 3-cycles are not very interesting, and so we generate permutations
that do not contain these cycles.
transposition tight permutations
We generate these permutations by randomly partitioning {b1 , . . . , bn+1 } into sets of
size three, such that no set contains bi and bi+1 for any i. Each of these partitions
represents a transposition that acts on the three black edges in the set. We then
apply a sequence of these (n + 1)/3 transpositions, in random order, to the identity
permutation. The resulting permutation has a cycle graph that contains only 3-cycles
such that no 3-cycle contains bi and bi+1 , for some i. Furthermore, we know that the
permutation is transposition tight. (This is in contrast to 3-cycle permutations which
may or may not be transposition tight.) Note, however, that we cannot claim that a
Chapter 3. Sorting by Transpositions
92
permutation generated in this way is chosen uniformly at random from the set of all
such permutations. Note also that these permutations can again only be generated if
n ≡ 2 (mod 3).
3.6.3
How the algorithms performed
Each algorithm was tested with each kind of permutation against 100 permutations of
sizes 8, 17, 26, 53, 107, 431 and 1727. (Remember that n + 1 must be divisible by three
in order for it to be possible to decompose a cycle graph into 3-cycles only.) Some tests
were not performed because the space or time requirements of the algorithms were too
great.
The algorithm branch was allowed to search for 15 minutes of processing time. The
algorithm rpt random was allowed to search for 5 minutes of processing time. All the
algorithms were written in Ada 95, and compiled with the Gnat Ada 95 compiler. All
the tests were performed on a Sun SPARCstation-20.
The performance of the various algorithms is presented in tables later in this section.
Recall that, for a given permutation, each algorithm (except lower bound) returns a
sequence of transpositions that sorts the permutation (a sorting sequence). We now
describe the column headings of the tables. Of course, for lower bound the tables
present figures derived from values calculated as the lower bound, rather than from the
length of sorting sequence obtained.
Average
The average length of sorting sequence found by the algorithm over the 100 instances.
Min
The shortest length of sorting sequence found by the algorithm.
Max
The longest length of sorting sequence found by the algorithm.
Av gap
The average difference between the length of the sorting sequence found by the algorithm and the lower bound of Theorem 3.5.3 over the 100 instances.
Max gap
The biggest difference between the length of the sorting sequence found by the algorithm and the lower bound of Theorem 3.5.3.
Chapter 3. Sorting by Transpositions
93
Match lb
The number of times the length of the sorting sequence found by the algorithm was
equal to the lower bound of Theorem 3.5.3.
Exact
The number of times the length the sorting sequence found by the algorithm was equal
the exact transposition distance of the permutation (when the exact transposition
distance of the permutation is known).
Av Time
The average time in seconds required by the algorithm. Note that for branch and
rpt random this time excludes the time required to establish initial upper bound using
alt approx.
For the algorithm branch there are two more table headings, they are:
Finished
The number of times the algorithm completed its search inside the time allowed.
Longest
The longest time (in seconds) required by the algorithm to complete its search, when
the algorithm did complete its search.
94
Chapter 3. Sorting by Transpositions
n
Average
Min
Max
8
17
26
53
107
431
1727
3.96
8.43
12.75
26.18
53.07
214.67
862.45
3
7
12
25
52
213
860
4
9
13
27
54
216
864
Table 3.1: random permutations, lower bound
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
4.20
8.75
13.02
26.51
53.34
214.96
862.83
3
7
12
25
52
213
860
5
10
14
28
55
217
865
0.24
0.32
0.27
0.33
0.27
0.29
0.38
1
2
1
2
1
1
2
76
69
73
68
73
71
63
98
88
73
68
73
71
63
0.0
0.0
0.0
0.0
0.0
0.2
3.2
Table 3.2: random permutations, approx
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
4.20
8.75
13.00
26.54
53.36
214.98
862.70
3
7
12
25
52
213
860
5
10
14
28
55
217
865
0.24
0.32
0.25
0.36
0.29
0.31
0.25
1
2
1
1
1
2
2
76
69
75
64
71
70
76
98
88
75
64
71
70
76
0.0
0.0
0.0
0.0
0.0
0.2
3.3
Table 3.3: random permutations, alt approx
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
4.36
10.01
16.12
3
8
14
5
12
19
0.40
1.58
3.37
1
5
7
60
17
0
82
20
0
0.0
2.3
30.2
Table 3.4: random permutations, tlis
95
Chapter 3. Sorting by Transpositions
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
4.26
9.04
13.34
27.14
54.28
215.93
863.72
3
7
12
25
52
213
861
5
11
15
30
58
219
867
0.30
0.61
0.59
0.96
1.21
1.26
1.27
1
2
2
3
4
4
3
70
44
47
31
23
16
27
92
62
47
31
23
16
27
0.0
0.0
0.0
0.0
0.2
8.2
514.6
Table 3.5: random permutations, random
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
4.18
8.64
12.90
26.29
53.13
214.70
862.61
3
7
12
25
52
213
860
5
9
14
27
54
216
865
0.22
0.21
0.15
0.11
0.06
0.03
0.16
1
2
1
1
1
1
1
78
80
85
89
94
97
84
100
99
85
89
94
97
84
66.0
60.0
45.0
33.0
18.2
19.9
259.8
Table 3.6: random permutations, rpt random
n
Average
Min
Max
Av gap
Max gap
Match lb
Finished
Exact
Av time
Longest
8
17
26
53
107
431
1727
4.18
8.63
12.91
26.29
53.13
214.70
862.45
3
7
12
25
52
213
860
5
9
14
27
54
216
864
0.22
0.20
0.16
0.11
0.06
0.03
0.00
1
1
1
1
1
1
0
78
80
84
89
94
97
100
100
100
84
89
94
97
100
100
100
84
89
94
97
100
0.0
23.1
144.4
99.0
54.0
27.1
2.4
0.0
697.5
42.6
0.2
0.2
1.1
66.0
Table 3.7: random permutations, branch
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
8
17
26
53
107
431
1727
4.18
8.63
12.90
26.29
53.13
214.70
862.45
3
7
12
25
52
213
860
5
9
14
27
54
216
864
0.22
0.20
0.15
0.11
0.06
0.03
0.00
1
1
1
1
1
1
0
78
80
85
89
94
97
100
100
100
85
89
94
97
100
Table 3.8: random permutations, best
96
Chapter 3. Sorting by Transpositions
n
Average
Min
Max
8
17
26
53
107
431
1727
3.19
6.05
9.00
18.00
36.00
144.00
576.00
3
6
9
18
36
144
576
4
7
9
18
36
144
576
Table 3.9: 3-cycle permutations, lower bound
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
3.32
6.57
9.78
19.21
37.18
145.12
577.13
3
6
9
18
36
144
576
4
8
11
22
39
148
580
0.13
0.52
0.78
1.21
1.18
1.12
1.13
1
2
2
4
3
4
4
87
53
37
20
24
21
24
100
92
86
22
24
21
24
0.0
0.0
0.0
0.0
0.0
0.2
2.9
Table 3.10: 3-cycle permutations, approx
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
3.32
6.57
9.78
19.21
37.18
145.12
577.13
3
6
9
18
36
144
576
4
8
11
22
39
148
580
0.13
0.52
0.78
1.21
1.18
1.12
1.13
1
2
2
4
3
4
4
87
53
37
20
24
21
24
100
92
86
22
24
21
24
0.0
0.0
0.0
0.0
0.0
0.2
2.9
Table 3.11: 3-cycle permutations, alt approx
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
4.35
9.92
15.87
3
6
11
6
13
18
1.16
3.87
6.87
2
6
9
25
1
0
25
1
0
0.0
2.2
27.8
Table 3.12: 3-cycle permutations, tlis
97
Chapter 3. Sorting by Transpositions
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
3.32
6.57
9.85
19.07
37.23
145.24
577.42
3
6
9
18
36
144
576
4
8
12
22
40
147
579
0.13
0.52
0.85
1.07
1.23
1.24
1.42
1
2
3
4
4
3
3
87
53
33
30
21
20
11
100
92
83
32
21
20
11
0.0
0.0
0.0
0.0
0.0
0.2
3.5
Table 3.13: 3-cycle permutations, random
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
3.32
6.54
9.69
18.44
36.55
144.53
576.54
3
6
9
18
36
144
576
4
8
11
20
37
145
577
0.13
0.49
0.69
0.44
0.55
0.53
0.54
1
2
2
2
1
1
1
87
55
41
57
45
47
46
100
94
92
61
45
47
46
39.0
135.0
177.0
129.1
165.0
159.2
166.3
Table 3.14: 3-cycle permutations, rpt random
n
Average
Min
Max
Av gap
Max gap
Match lb
Finished
Exact
Av time
Longest
8
17
26
53
107
431
3.32
6.48
9.61
18.47
36.69
144.59
3
6
9
18
36
144
4
8
11
20
39
145
0.13
0.43
0.61
0.47
0.69
0.59
1
1
2
2
3
1
87
57
41
54
39
41
100
100
100
54
39
41
100
100
100
59
39
41
0.0
0.0
8.1
419.0
550.5
538.4
0.0
1.0
630.5
262.4
83.7
278.5
Table 3.15: 3-cycle permutations, branch
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
8
17
26
53
107
431
1727
3.32
6.48
9.61
18.43
36.55
144.53
576.54
3
6
9
18
36
144
576
4
8
11
19
37
145
577
0.13
0.43
0.61
0.43
0.55
0.53
0.54
1
1
2
1
1
1
1
87
57
41
57
45
47
46
100
100
100
62
45
47
46
Table 3.16: 3-cycle permutations, best
98
Chapter 3. Sorting by Transpositions
n
Average
Min
Max
8
17
26
53
107
431
1727
3.00
6.00
9.00
18.00
36.00
144.00
576.00
3
6
9
18
36
144
576
3
6
9
18
36
144
576
Table 3.17: transposition tight permutations, lower bound
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
3.00
6.00
9.02
18.26
36.59
145.02
577.16
3
6
9
18
36
144
576
3
6
11
20
39
148
580
0.00
0.00
0.02
0.26
0.59
1.02
1.16
0
0
2
2
3
4
4
100
100
99
85
69
47
40
100
100
99
85
69
47
40
0.0
0.0
0.0
0.0
0.0
0.2
2.8
Table 3.18: transposition tight permutations, approx
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
3.00
6.00
9.02
18.26
36.59
145.02
577.16
3
6
9
18
36
144
576
3
6
11
20
39
148
580
0.00
0.00
0.02
0.26
0.59
1.02
1.16
0
0
2
2
3
4
4
100
100
99
85
69
47
40
100
100
99
85
69
47
40
0.0
0.0
0.0
0.0
0.0
0.2
2.8
Table 3.19: transposition tight permutations, alt approx
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
3.46
8.18
13.98
3
6
10
5
11
17
0.46
2.18
4.98
2
5
8
55
7
0
55
7
0
0.0
1.7
23.4
Table 3.20: transposition tight permutations, tlis
99
Chapter 3. Sorting by Transpositions
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
3.00
6.00
9.02
18.20
36.69
145.01
577.13
3
6
9
18
36
144
576
3
6
11
20
38
148
580
0.00
0.00
0.02
0.20
0.69
1.01
1.13
0
0
2
2
2
4
4
100
100
99
87
62
45
42
100
100
99
87
62
45
42
0.0
0.0
0.0
0.0
0.0
0.2
3.4
Table 3.21: transposition tight permutations, random
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
Av time
8
17
26
53
107
431
1727
3.00
6.00
9.00
18.00
36.00
144.00
576.00
3
6
9
18
36
144
576
3
6
9
18
36
144
576
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0
0
0
0
0
0
0
100
100
100
100
100
100
100
100
100
100
100
100
100
100
0.0
0.0
0.0
0.0
0.0
0.2
5.4
Table 3.22: transposition tight permutations, rpt random
n
Average
Min
Max
Av gap
Max gap
Match lb
Finished
Exact
Av time
Longest
8
17
26
53
107
431
3.00
6.00
9.00
18.01
36.11
144.20
3
6
9
18
36
144
3
6
9
19
37
146
0.00
0.00
0.00
0.01
0.11
0.20
0
0
0
1
1
2
100
100
100
99
89
81
100
100
100
99
89
81
100
100
100
99
89
81
0.0
0.0
0.0
11.7
102.3
195.7
0.0
0.0
0.0
264.3
118.5
713.1
Table 3.23: transposition tight permutations, branch
n
Average
Min
Max
Av gap
Max gap
Match lb
Exact
8
17
26
53
107
431
1727
3.00
6.00
9.00
18.00
36.00
144.00
576.00
3
6
9
18
36
144
576
3
6
9
18
36
144
576
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0
0
0
0
0
0
0
100
100
100
100
100
100
100
100
100
100
100
100
100
100
Table 3.24: transposition tight permutations, best
Chapter 3. Sorting by Transpositions
3.6.4
100
Analysis of the performance of the algorithms
The most important conclusions from these experiments are:
• for most permutations π, td(π) is equal to or very close to tl(π).
• for most randomly generated permutations π, we can evaluate td(π) and find an
optimal length sorting sequence.
We now review and summarise the performances of the various algorithms. First of all
we describe features of the algorithms that were noticeable in the experiment results.
Later we discuss general features of the experiment results.
approx and alt approx
The algorithms approx and alt approx have almost identical behaviour in all the
tests (Note that for 3-cycles permutations, and transposition tight permutations both
algorithms actually do behave in an identical manner, because there are no even cycles),
except that alt approx performs marginally better for larger values of n. It is because
of this slight advantage that alt approx was chosen as the upper bound used in branch.
We conclude that both algorithms contain good strategies for finding sorting sequences, since both algorithms find shorter sorting sequences, on average, than random.
It is perhaps a little surprising that the performance of approx is comparable to
alt approx when alt approx keeps 2-transpositions on even cycles up its sleeve in
case it gets stuck with the odd cycles. Might approx not waste its 2-transpositions
on even cycles, and get stuck looking for a 2-transposition more often? Perhaps these
algorithms have similar performance because approx hardly ever gets stuck for a 2transposition to apply, so it does not need to hold back 2-transpositions on even cycles
for emergencies.
Another reason for the similar performance is that the following lemma tells us that
after applying a 2-transposition that acts on an even cycle, we must be able to apply
a 2-transposition on the resulting permutation.
Lemma 3.6.1 If tG(π) contains an oriented even cycle C, then it is possible to apply
a sequence of two 2-transpositions on π.
Proof Since C is oriented, it must be fully oriented, so a 2-transposition exists that
splits C into two odd cycles and an even cycle. Now by Lemma 3.2.5 a 2-transposition
can be applied on the resulting permutation, because it contains an even cycle in its
cycle graph. 2
Therefore approx is unlikely to waste 2-transpositions on even cycles, since any
2-transposition on an even cycle must lead to at least one further 2-transposition.
Chapter 3. Sorting by Transpositions
101
Though neither algorithm achieves an approximation bound, each typically uses
only a handful of 0-transpositions to sort a given permutation, and far fewer 0-transpositions than the maximum allowed in order to achieve a 3/2-approximation bound
for transposition distance.
The algorithms are reasonably fast for all the permutations tested. The observed
run-time complexity was O(n2 ) for both algorithms.
tlis
The performance of tlis is clearly not as good as approx or alt approx. On average
it produces longer length sequences of transpositions to sort the given permutations,
and it requires much more time.
The results of these experiments would appear to refute Heath and Vergara’s claim
that this algorithm “often produces near-optimal results” [HV97]. As n increases the
number of permutations for which this algorithm finds an exact solution rapidly decreases. In fact, when n = 26 the algorithm fails to find an exact solution for any of
the cases generated of the transposition tight permutations which, as we will explain
later, we expect to be particularly easy to solve exactly.
Note that the other algorithms described by Guyer, Heath, and Vergara [GHV95],
which are based on repeatedly selecting a strip from the permutation and moving it
by a transposition so as to remove at least one breakpoint, require approximately n
transpositions to sort a random permutation, whereas approx and alt approx appear
to require approximately n/2 transpositions. In fact, as a result of Theorem 3.2.2,
we know that strips are an unimportant feature of a permutation, with respect to its
transposition distance.
The observed run-time complexity was O(n5 log n) as expected. In fact this means
that the algorithm requires too much time to solve the instances when n was larger
than 26.
random and rpt random
The algorithm random produces sequences of transpositions that are longer, on average,
than those produced by approx or alt approx. This is viewed as evidence that the
strategies employed by approx and alt approx are good strategies.
However, the algorithm rpt random finds sequences that are shorter on average
than those produced by approx or alt approx. In fact, for 3-cycle permutations and
transposition tight permutations this algorithm finds sorting sequences that are shorter,
on average, than all the other algorithms. For the random permutations rpt random
has similar performance to branch, except for larger values of n when, perhaps, counting the 2-transpositions takes so long that only a few iterations of random can be
performed inside the time limit.
Chapter 3. Sorting by Transpositions
102
The observed run-time complexity of random was O(n3 ) for the random permutations, and O(n2 ) for the other two types of permutations. Note that when the cycle
graph contains only 3-cycles, the method used to count 2-transpositions on π requires
only O(n) steps so the algorithm has O(n2 ) overall time-complexity. Note that, if n is
much larger than 1727 there are too many 2-transpositions on the random permutations
to be able to count them in a reasonable amount of time.
branch
For random permutations branch finds the shortest sorting sequences, on average. It
is worth noting that for the random permutations tested, when n ≥ 53 the algorithm
finds an exact sorting sequence in about a second, if it can determine an exact sorting
sequence. Note, of course, that this timing excludes the first call of alt approx.
For n = 1727 the algorithm ran out of space when attempting to solve the 3-cycle
permutation and transposition tight permutation instances. Therefore the tables do
not contain any figures for these instances.
best
A strange property in the tables for the hypothetical algorithm best on the random
permutations is the way that the column Exact starts with 100, then decreases before
increasing to near 100 again. We believe that three factors have created this behaviour.
The first factor is that, in some sense, proving td(π) = tl(π) is easier than proving
td(π) > tl(π). This is because if td(π) = tl(π) then we only need to find one sequence
of length tl(π) that sorts π. However if td(π) > tl(π) then we must show that all
sequences of length tl(π) fail to sort π, and this may involve a great deal of work.
The second factor is that we believe that if td(π) = tl(π) then it is likely that there
will be many sequences of this length that sort π, which makes it easier to find such a
sorting sequence. The evidence that supports this conjecture is the good performance of
the randomised algorithm rpt random, and the approximation algorithms alt approx
and approx.
The third factor is that, for random permutations, it would appear that as n increases, the lower bound for transposition distance becomes a more accurate measure of
exact transposition distance. That is, as n increases, it is more likely that td(π) = tl(π).
So for small values of n exhaustive search is enough to find the exact transposition
distance of all the permutations given. Then for slightly larger values of n, exhaustive
search is unable to determine the exact transposition distance of permutations that
have td(π) > tl(π) in a reasonable amount of time. For much larger values of n, it
is very likely that td(π) = tl(π) and that branch or rpt random will quickly find a
sequence that proves it.
We summarise this behaviour in Figure 3.24. The white ribbon in the graph represents the number of permutations for which the lower bound was demonstrated to
103
Chapter 3. Sorting by Transpositions
100
90
80
70
%
60
50
40
30
20
10
0
best
8
17
search
26
53
n
lb
107
431
1727
Figure 3.24: Performance of best.
be the exact distance. The grey ribbon is an estimate of the percentage of permutations of length n, that have td(π) > tl(π), for which exhaustive search can prove
td(π) > tl(π) in a reasonable amount of time. The black ribbon is the maximum of
the other two edges, and represents the number of permutations for which the exact
distance is known.
For all the permutations tested, the length of the sorting sequence found by best
was at most three greater than the lower bound of Theorem 3.5.3. This suggests that,
in practice, this lower bound is a very accurate lower bound.
For all the permutations tested, the length of sorting sequence found by best was
never greater than bn/2c + 1. We view this as supporting evidence for Conjecture
3.5.1. If the transposition diameter were to be close to 3n/4, the tightest known upper
bound for tD(π), then we would expect to have found permutations with transposition
distance noticeably greater than bn/2c + 1.
For random permutations the minimum distance and maximum distance found by
best are remarkably close to the average distance, which itself is remarkably close to
dn/2e. Perhaps it may be possible to show that the expected transposition distance of
a random permutation is dn/2e.
Chapter 3. Sorting by Transpositions
104
3 18 12 17 21 25 8 23 2 13 10 19 9 20 5 7 4 11 24 16 14 6 1 15 22
3 18 12 17 21 25 8 23 2 13 10 19 9 20 5 6 7 4 11 24 16 14 1 15 22
1 3 18 12 17 21 25 8 23 2 13 10 19 9 20 5 6 7 4 11 24 16 14 15 22
1 2 13 10 19 9 20 5 6 7 4 11 3 18 12 17 21 25 8 23 24 16 14 15 22
1 2 3 18 13 10 19 9 20 5 6 7 4 11 12 17 21 25 8 23 24 16 14 15 22
1 2 3 4 11 12 17 21 25 18 13 10 19 9 20 5 6 7 8 23 24 16 14 15 22
1 2 3 4 11 12 17 18 13 10 19 9 20 21 25 5 6 7 8 23 24 16 14 15 22
1 2 3 4 5 6 7 8 23 24 16 14 15 22 11 12 17 18 13 10 19 9 20 21 25
1 2 3 4 5 6 7 8 9 23 24 16 14 15 22 11 12 17 18 13 10 19 20 21 25
1 2 3 4 5 6 7 8 9 23 24 16 17 18 13 14 15 22 11 12 10 19 20 21 25
1 2 3 4 5 6 7 8 9 13 14 15 22 11 12 10 23 24 16 17 18 19 20 21 25
1 2 3 4 5 6 7 8 9 13 14 15 16 17 18 19 20 21 22 11 12 10 23 24 25
1 2 3 4 5 6 7 8 9 11 12 10 13 14 15 16 17 18 19 20 21 22 23 24 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Figure 3.25: A sequence of 13 transpositions that sorts π.
Discussion
The algorithm rpt random very quickly finds an exact sequence of transpositions to
sort each of the transposition tight permutations. These permutations are generated
by applying dn/3e non-interfering transpositions on π. Therefore, a transposition tight
permutation for which there are many optimal length sorting sequences is more likely to
be generated by this method than a transposition tight permutation for which there are
relatively few optimal length sorting sequences. So we cannot conclude that rpt random
is a good test for determining if a permutation is transposition tight, because there may
be some permutations that were not generated in our tests, for which it is much more
difficult to find an exact sequence of transpositions.
However, the algorithm rpt random does perform much better with these permutations than the other algorithms, for all of which the number of permutations for which
the exact solution is found decreases as n increases.
Unfortunately, there are some permutations with n as small as 25 for which our
algorithms are unable to determine the exact transposition distance. For example,
π = [3 18 12 17 21 25 8 23 2 13 10 19 9 20 5 7 4 11 24 16 14 6 1 15 22] has tl(π) = 12.
The algorithm approx finds a sorting sequence, shown in Figure 3.25, of length 13.
However a week’s computation by algorithms branch and rpt random has not been
able to establish if td(π) = 12, or td(π) = 13.
Chapter 3. Sorting by Transpositions
105
In general, it would appear that, when td(π) = tl(π), then there are many sequences
of length tl(π) that sort π. Therefore we conjecture that td(π) = 13, and that the exact
algorithms run out of time before proving this. However, it is possible that there are
relatively few sorting sequences of length 12, and that is why the algorithms have been
unable to find one of these sequences.
It is possible to discard some transpositions during the search because we know that
they would lead to a permutation that we have already searched from. For instance for
many transpositions τ1 and τ2 , π · τ1 · τ2 = π · τ2 · τ1 . A branch and bound algorithm was
implemented that discarded many possible search paths for this reason. Certainly this
alternative algorithm detects permutations that have td(π) > tl(π) much more quickly
than the standard algorithm. However, there appear to be very few permutations like
this, and in tests the alternative algorithm found hardly any new permutations like
this.
It is also possible to change the branch and bound algorithm by removing the use
of alt approx from inside the search procedure of branch. This alteration to the
algorithm speeds up the detection of permutations that have td(π) > tl(π), but slows
down the detection of permutations that have td(π) = tl(π). Since most permutations
have td(π) = tl(π), this algorithm finds the exact distance of fewer permutations than
the standard algorithm, when given the same time constraints.
We also tested all the algorithms with so called m-permutations. These permutations are generated by applying a sequence of m random transpositions to the identity
permutation. We generated m-permutations of various sizes with m = dn/2e, dn/3e,
and dn/4e. As with the other tests, the algorithms branch and rpt random found the
shortest sorting sequences on average. These algorithms could determine the exact
transposition distance of about 90 percent of these permutations. Interestingly td(π)
was noticeably smaller than m, on average.
3.7
Conclusion and open problems
We conclude this chapter with some conjectures and open problems. We begin with a
simple conjecture that would allow us to disregard (−2)-transpositions.
Conjecture 3.7.1 For any permutation π, there is a sequence of transpositions of
length td(π) that sorts π and contains no (−2)-transpositions.
We make this conjecture because applying a (−2)-transpositions would appear to
be a negative step towards sorting π, however we have no insight as to how to prove
this conjecture.
We may chip away at the problem of sorting by transposition, by cataloguing a
list of classes of permutations for which the exact transposition distance is known.
Chapter 3. Sorting by Transpositions
106
Here is a list of permutations for which the exact transposition distance is known
(of course, permutations transposition equivalent to the following permutations have
known transposition distance too):
• the reverse permutation Rn (Section 3.5.3).
• the permutation ωr (Section 3.3.4).
• permutations with only 2-cycles in their cycle graphs. We can sort such permutations by repeated application of the 2-transpositions described in the proof of
Lemma 3.4.8.
• the permutation πn = [2 4 . . . n 1 3 . . . n − 1] (where n is even). Such a
permutation has only one cycle in its cycle graph, therefore by Theorem 3.2.3,
td(π) ≥ n/2. It is clear that this bound can be achieved by a sequence of n/2
transpositions that each move one odd element into the correct place in the
initial increasing sequence. Note that a permutation formed by concatenating
two increasing sequences is transposition equivalent to πn , for some n.
It is an open problem to extend this list. Of course a polynomial-time algorithm for
sorting by transpositions would make this list redundant, and proving whether sorting
by transpositions is in P remains an open problem. It may be possible to find a hard
special case of sorting by transpositions. The problem of deciding if a permutation is
transposition tight (TTDP) may be a useful problem for resolving whether sorting by
transpositions is NP-hard.
A permutation must have a cycle graph containing only 3-cycles if it is to be transposition tight, and at least one cycle in each component of the overlap graph must be
oriented. However, we know little more about the problem. The way that 3-cycles can
interact with each other is very rich and varied. Deciding if a permutation is transposition tight seems to be much more complicated than deciding if a permutation is reversal
tight. The experiments show that the algorithm rpt random could recognise the permutations that were generated to be transposition tight. However, as was explained
before, the method used to generate these permutations is more likely to generate a
permutation that has many optimal length sorting sequences, than a permutation with
relatively few optimal length sorting sequences.
The approximation algorithms of Section 3.6 appear to perform least effectively
when given input permutations containing only 3-cycles in their cycle graphs. So an
algorithm to solve TTDP, or some good heuristics for the problem would be useful to
help improve our approximation algorithm.
Chapter 4
Sorting by Block-Interchanges
4.1
Introduction
In this chapter we introduce an operation, the block-interchange, in which two substrings, or blocks, are swapped in a permutation. We demonstrate a polynomial-time
algorithm for calculating the block-interchange distance of a permutation (i.e. the minimum number of block-interchanges required to transform the permutation to the identity). We also determine the block-interchange diameter of the symmetric group.
A block-interchange can be viewed as a generalisation of a transposition. In a
block-interchange two non-intersecting substrings of any length are swapped in the
permutation. In a transposition the substrings must be adjacent.
A block-interchange β = β(i, j, k, l), where 1 ≤ i < j ≤ k < l ≤ n + 1, is applied
to π by exchanging the blocks [π(i) . . . π(j − 1)] and [π(k) . . . π(l − 1)]. So, if
j 6= k then the resulting permutation π · β = [π(0) . . . π(i − 1) π(k) . . . π(l −
1) π(j) . . . π(k − 1) π(i) . . . π(j − 1) π(l) . . . π(n + 1)], and if j = k then π · β =
[π(0) . . . π(i − 1) π(k) . . . π(l − 1) π(i) . . . π(j − 1) π(l) . . . π(n + 1)]. (As is standard,
we assume π(0) = 0 and π(n+1) = n+1.) Note that when j = k the block-interchange
is also a transposition.
The block-interchange distance of π, bd(π), is the length of a shortest sequence
of block-interchanges that transforms π into the identity permutation. For example
the permutation π = [5 2 4 1 3] has block-interchange distance 2, as demonstrated
in Figure 4.1. Sorting by block-interchanges is the problem of finding a sequence of
block-interchanges of length bd(π) that sorts π.
We find transposition breakpoints, as defined in Chapter 3, to be useful in studying
the problem of sorting by block-interchanges. To recall, a transposition breakpoint is
a value of i such that 1 ≤ i ≤ n + 1 and π(i) − π(i − 1) 6= 1. In the rest of this chapter,
for conciseness, we talk of breakpoints where we mean transposition breakpoints.
The number of breakpoints in the permutation π is denoted by tb(π). Note that
the identity permutation is the only permutation with no breakpoints, and a blockinterchange can change tb(π) by at most 4.
107
108
Chapter 4. Sorting by Block-Interchanges
5
2
4
1
3
1
3
2
4
5
1
2
3
4
5
Figure 4.1: An example of sorting by block-interchanges
In this chapter we describe a polynomial-time algorithm for evaluating bd(π). This
is achieved by first demonstrating that for all permutations, except the identity permutation, a block-interchange exists that removes at least two breakpoints. Then
it is proved using a graph model for the permutations that an algorithm which repeatedly applies these block-interchanges sorts π using as few block-interchanges as
is possible. We then study the permutations of n elements which require the most
block-interchanges in order to sort them and show that the block-interchange diameter
of the symmetric group Sn is bn/2c.
4.2
Minimal block-interchanges
Lemma 4.2.1 It is always possible to find a block-interchange that removes at least
two breakpoints from a given permutation, π, unless π is the identity permutation.
Proof Since π is not the identity permutation, there must be at least two elements
in π that appear in the wrong order, i.e., there must be an x and a y such that x < y
but π = [. . . y . . . x . . .].
Now choose x to be the smallest such value and choose y to be the largest value in
π to the left of this x. Then x − 1 must be to the left of y in π since otherwise this
would contradict the choice of x. Similarly y + 1 must be to the right of x in π. Note
that in fact x is the smallest value such that π(x) 6= x. Hence π has the form
π = [1 . . . x − 1 . . . y . . . x . . . y + 1 . . .].
The block-interchange β = β(π −1 (x − 1) + 1, π −1 (y) + 1, π −1 (x), π −1 (y + 1)) transforms π into π · β:
π · β = [1 . . . x − 1 x . . . . . . . . . y y + 1 . . .].
Notice that before this transformation there were breakpoints at each place where
the block-interchange cut π. After the block-interchange there are at least two fewer
breakpoints in the permutation. Hence the lemma. 2
We call the block-interchange described in the above lemma the minimal block
interchange. Since we can remove at most four breakpoints with any block-interchange
109
Chapter 4. Sorting by Block-Interchanges
a
b
c
d
e
f
g
h
a
f
g
d
e
b
c
h
(a)
a
b
c
d
e
f
a
d
e
b
c
f
(b)
Figure 4.2: The black edges that change when a block-interchange is applied: (a) when
the block-interchange is not a transposition, and (b) when the block-interchange is a
transposition.
and a minimal block-interchange removes at least two breakpoints, then an algorithm
that repeatedly applies the minimal block-interchanges is a 2-approximation algorithm
for block-interchange distance. However when we examine these block-interchanges on
a graph model we discover that, in fact, this algorithm is an exact algorithm.
4.3
The graph model
The graph model we use is that of the transposition cycle graph as presented in Chapter
3. The transposition cycle graph, tG(π), of π is a directed edge-coloured graph with
vertex set {0, . . . , n + 1}, grey edge set {(i, i + 1) : 0 ≤ i ≤ n}, and black edge set
{(π(i), π(i − 1)) : 1 ≤ i ≤ n + 1}. Note that the edge (u, v) is directed from u to v.
Since every incoming edge of a vertex can be uniquely paired with an outgoing edge
of the other colour, we can easily completely decompose the graph into alternating
cycles, and furthermore this decomposition is unique. The identity permutation is
the only permutation that has n + 1 alternating cycles in its graph. The number of
alternating cycles in tG(π) is denoted by tc(π).
A block-interchange changes only black edges in the graph. Three or four black
edges are removed from the graph and replaced by new black edges as shown in Figure
4.2.
Lemma 4.3.1 When a minimal block-interchange is applied tc(π) increases by two.
Chapter 4. Sorting by Block-Interchanges
110
Proof A minimal block interchange may change 3 or 4 black edges of tG(π). If it
changes 3 edges then those edges must be part of one cycle. If it changes 4 edges then
those edges must be part of either one or two cycles. The three possible cases are shown
in Figure 4.3 (a), (b) and (c). In each of these cases tc(π) increases by two. Note that
in the figure, paths which start and finish with a grey edge are represented by dotted
grey edges. 2
Lemma 4.3.2 It is impossible to increase tc(π) by more than two with a single blockinterchange.
Proof A block interchange changes at most four black edges in the graph, so the only
way that we could increase tc(π) by more than two with a single block-interchange
would be if one big cycle was broken into four cycles, but this is impossible. A block
interchange that changes black edges from four different cycles results in two cycles.
By symmetry, a block interchange that results in four cycles must have acted on black
edges from two cycles (an example of this is shown in Figure 4.3 (c)). So tc(π) can be
increased by at most two with a single block-interchange. 2
4.4
Block-interchange distance
We can now present the main result of this chapter.
Theorem 4.4.1 The block-interchange distance bd(π) of a permutation π is
bd(π) =
(n + 1) − tc(π)
2
where tc(π) is the number of alternating cycles in the cycle graph of π.
Proof By Lemma 4.3.1, bd(π) ≤ (tc(ın ) − tc(π))/2, and by Lemma 4.3.2, bd(π) ≥
(tc(ın ) − tc(π))/2, where ın is the identity permutation of length n. So bd(π) = ((n +
1) − tc(π))/2, since tc(ın ) = n + 1. 2
An algorithm that calculates the block-interchange distance is shown in Figure 4.4.
The algorithm performs a depth-first search of tG(π), counting the number of cycles,
and hence runs in linear time.
An algorithm that performs an optimal sequence of block-interchanges is shown in
Figure 4.5. This algorithm has O(n2 ) complexity since all the steps inside the while
loop can be executed in linear time, and the loop itself is executed bn/2c times at most.
4.5
Block-interchange diameter
The block-interchange diameter, bD(n), of the symmetric group Sn , is the maximum
value of bd(π) taken over all n-element permutations. Rn is the reverse permutation
111
Chapter 4. Sorting by Block-Interchanges
x−1
a
y
x
b
y+1
x−1
x
b
a
y
y+1
(a)
x−1
a
y
b
c
x
d
y+1
x−1
x
d
b
c
a
y
y+1
(b)
x−1
a
y
b
c
x
d
y+1
x−1
x
d
b
c
a
y
y+1
(c)
Figure 4.3: How a minimal block-interchange changes tG(π).
Chapter 4. Sorting by Block-Interchanges
112
algorithm bd(π: Permutation) is
visited: array of Boolean;
j, cycles: Integer;
begin
initialise all components of visited as f alse;
cycles:= 0;
for i in 0 .. |π| loop
if not visited(i) then
cycles:= cycles+1;
j:= i;
while not visited(j) loop
visited(j):= true;
j:= j + 1; — follow grey edge
j:= π(π −1 (j) − 1); — follow black edge
end loop;
end if ;
end loop;
return (n + 1 − cycles)/2;
end bd;
Figure 4.4: An algorithm to calculate bd(π).
algorithm bd sequence(π: Permutation) is
x, y: Integer;
begin
while π 6= ı loop
locate the smallest x s.t. π(x) 6= x;
find the largest value, y, between x − 1 and x in π;
π:= π · β(x, π −1 (y) + 1, π −1 (x), π −1 (y + 1));
end loop;
end bd sequence;
Figure 4.5: An algorithm to perform an optimal sequence of block-interchanges.
113
Chapter 4. Sorting by Block-Interchanges
of length n, i.e., Rn = [n n − 1 . . . 1]. It is easy to show that tc(Rn ) is given by the
formula
(
2 if n is odd,
tc(Rn ) =
1 otherwise.
So tc(Rn ) is as small as possible (since tc(π) ≡ n + 1 (mod 2) for any permutation π
of length n). Hence bD(n) is achieved by the reverse permutation and is bn/2c.
There are many other permutations which achieve this upper bound, e.g., the permutation [2 4 6 8 1 3 5 7] contains only one cycle in its cycle graph, and the family of
permutations like it (that have even numbers in ascending order preceding odd numbers
in ascending order) have the same value for tc(π) as the reverse permutation. Table
4.1 shows how many permutations achieve the bD(n) for small values of n.
n
number
1
1
2
1
3
5
4
8
5
84
6
180
7
3044
8
8064
9
193248
10
604800
Table 4.1: The number of permutations of size n that achieve bD(n).
It is perhaps a little surprising that even though a block-interchange would appear
to be a much more powerful primitive operation for sorting than a transposition, the
block-interchange diameter (bn/2c) is only one less than the conjectured transposition
diameter (bn/2c + 1).
4.6
Conclusion
A block-interchange is a generalisation of a transposition. However, in contrast with
what is known for transpositions, we have derived a polynomial-time algorithm for
sorting by block-interchanges, and have also evaluated the block-interchange diameter
of Sn , The complexity of sorting by transpositions is unresolved, and the transposition
diameter of Sn is not known.
We can think of a block-interchange as an operation in which three adjacent substrings of π are rearranged in any order. This is an alternative generalisation of a
transposition, since a transposition, of course, rearranges two adjacent substrings of π
in the only different order possible. As far as we know, the problem of sorting π using
operations that rearrange k adjacent substrings of π in any order, has not been studied
for k ≥ 4.
Now that we have a polynomial-time algorithm for sorting by block-interchanges, we
might ask, what can we say about the problem if we allow the sequences being compared
to contain repeated characters? In Chapter 5 we study this and other similar problems.
In particular, we show that this generalised version of sorting by block-interchanges is
NP-hard.
Chapter 5
Sorting Strings by Global
Transformations
5.1
Introduction
The previous chapters were devoted to the problems of sorting permutations by reversals, transpositions and block-interchanges. Corresponding problems can be defined on
strings (that may contain repeated characters) instead of permutations. These problems, together with the problem on strings that corresponds to sorting permutations
by prefix-reversals, are investigated in this chapter.
It was shown in a previous chapter that, for permutations, transforming π into φ
is equivalent to transforming φ−1 · π into ı. However, there is no analogue of this result
for strings. Therefore the string problems are expressed in terms of two strings.
Given strings S and T : the reversal distance rd(S, T ) is the minimum number of
reversals required to transform S into T ; the prefix-reversal distance prd(S, T ) is the
minimum number of prefix-reversals required to transform S into T ; the transposition
distance td(S, T ) is the minimum number of transpositions required to transform S
into T ; and the block-interchange distance bd(S, T ) is the minimum number of blockinterchanges required to transform S into T . It is impossible to insert or delete characters using reversals, prefix-reversals, transpositions, or block-interchanges, so T must
be a rearrangement of S, since otherwise it would be impossible to transform S into
T . We say that S and T are related if T is a rearrangement of S.
For each of the four transformations, the distance problem on permutations is a
special case of the distance problem on strings. Since, for example, rd(π) = rd(π, ı n ).
Therefore, for each of the transformations, the distance problem on strings is at least as
hard as the sorting problem on permutations. Thus, finding reversal distance on strings
is NP-hard since sorting permutations by reversals is NP-hard [Cap97b]. Similarly, finding prefix-reversal distance is NP-hard since sorting permutations by prefix-reversals is
NP-hard [HS].
114
Chapter 5. Sorting Strings by Global Transformations
115
If the strings are drawn from a fixed size alphabet that is smaller than the length
of the strings, then the sorting problems on permutations are no longer special cases
of the distance problems on strings. In this chapter, distance problems on strings that
are drawn from a binary alphabet {0, 1} are studied. Of course, the distance problems
on a larger alphabet include the distance problems on a binary alphabet as a special
case. So the distance problems on a larger alphabet are at least as hard as the distance
problems on a binary alphabet.
The next section (Section 5.2) contains a few useful definitions. Sections follow for
each of the four distance problems. NP-completeness results are proved in Section 5.7
for two of the problems.
5.2
Definitions
A reversal, or prefix-reversal, on a string will be represented by enclosing in square
brackets the substring that has to be reversed. For example, 0[1010110]1 = 001101011.
A similar notation will be used to describe transpositions and block-interchanges,
though in both cases two substrings need to be bracketed.
Let 0k represent a string of zeros of length k, 1k represent a string of ones of length
k, and in general let S k represent the string obtained by concatenating k copies of S,
for any string S. Larger strings can be built from smaller strings by concatenating
them together using ++. For example, 02 ++ 13 = 00111. Concatenation (++) can also
P
be used in a similar way to summation ( ). For example, ++3i=1 0i 1 = 010010001. Note
that sometimes, as in the last example, the ++ symbol is omitted, and the strings to
be concatenated are simply placed side by side.
Let Bn be the set of binary strings of length n. For a particular transformation,
the diameter of Bn is the maximum distance between any two related binary strings
of length n. Define Bk = 0k ++ 1k , and Ck = (10)k . For example, B4 = 00001111 and
C4 = 10101010. These strings are particularly useful for establishing diameter results
in later sections.
Let S −1 represent the string derived from S by switching zeros and ones. Let S
represent the string S in reverse order. If S = 0100110001 then S −1 = 1011001110,
−1
S = 1000110010, and S = 0111001101.
Strings X and Y are isomorphic to strings S and T if:
(i)
(ii)
(iii)
(iv)
X
X
X
X
= S and Y
= S −1 and
= S and Y
−1
= S and
= T , or X = T
Y = T −1 , or X
= T , or X = T
−1
Y = T , or X
and Y = S, or
= T −1 and Y = S −1 , or
and Y = S, or
−1
−1
=T
and Y = S .
Obviously, if X and Y are isomorphic to S and T , then rd(X, Y ) = rd(S, T ), td(X, Y ) =
td(S, T ), and bd(X, Y ) = bd(S, T ).
Chapter 5. Sorting Strings by Global Transformations
116
Strings X and Y are prefix isomorphic to strings S and T if:
(i) X = S and Y = T , or X = T and Y = S, or
(ii) X = S −1 and Y = T −1 , or X = T −1 and Y = S −1 .
If X and Y are prefix isomorphic to S and T , then prd(X, Y ) = prd(S, T ).
Define lcp(S, T ) and lcs(S, T ) to be the lengths of the longest common prefix and
the longest common suffix, respectively, of S and T .
A block of zeros is a maximal length substring that consists only of the character
0 repeated. A block of ones is defined similarly. Let b(S) denote the total number
of blocks in S, and z(S) denote the number of blocks of zeros in S. So, for example,
b(001110101) = 6, and z(001110101) = 3.
In the sections that follow we will often know a prefix, a suffix and sometimes a
substring of a string, but not know (or care about) the rest of the string. In such cases
we represent the string by joining the known parts with dots (. . .). For example, if S
has prefix ‘01’, a substring ‘00’ and suffix ‘11’ then we write S = 01 . . . 00 . . . 11.
By contrast, sometimes we will need to represent strings that have a repeating
pattern. In such cases we will use a twiddle symbol (∼). For example, Ck = 1010 ∼ 10.
5.3
Reversal distance between binary strings
In this section, we present a lower bound and an upper bound for reversal distance
between binary strings. These bounds are then used to determine the reversal diameter
of Bn , and also to identify some strings that achieve the reversal diameter. A restricted
version of the problem, that is in some ways analogous to sorting permutations by
reversals, is shown to be solvable in polynomial-time. However, in Section 5.7 the
general problem of determining reversal distance between two strings is shown to be
NP-hard.
5.3.1
A lower bound
In this section the concept of breakpoint from permutation sorting problems is adapted
for use with string sorting problems. This new kind of breakpoint is then used to find
a lower bound for reversal distance. Remember that, for permutations, two elements
form a breakpoint if they are adjacent in π but not adjacent in the identity permutation.
Substrings of length 2 represent adjacencies in strings S and T , so our definition of
breakpoints on strings will be based on these substrings.
If S contains more ‘00’ substrings than T , then each extra ‘00’ must be broken, by
a reversal, at some time in the transformation from S into T . Similarly if T contains
more ‘00’ substrings than S, then each extra ‘00’ must be created by a reversal. Each
extra ‘00’ in S or T is a reversal breakpoint. An obvious difference between breakpoints
on strings and on permutations is that, on strings, the specific location of a breakpoint
may not necessarily be identified. For instance, if S contains three ‘00’ substrings and
Chapter 5. Sorting Strings by Global Transformations
117
T contains only two ‘00’ substrings, then one of the ‘00’ substrings in S is a breakpoint,
but no particular ‘00’ substring of S is selected as the breakpoint. Breakpoints occur
with ‘01’, ‘10’, and ‘11’ substrings as well. However, care must be taken, because
reversals can convert ‘01’ substrings into ‘10’ substrings. Therefore, ‘01’ substrings
and ‘10’ substrings are counted together when considering reversal breakpoints.
Breakpoints can also be contributed from the beginning and end of both strings.
For example, if S(1) 6= T (1), then position one contributes breakpoints. In order to
deal with these breakpoints, S and T are extended by adding special characters α at
the beginning, and ω at the end of both strings. These breakpoints can then be counted
by comparing the number of occurrences of the substrings ‘α0’, ‘α1’, ‘0ω’, and ‘1ω’ in
both strings. Adding α and ω to S and T , is similar to adding 0 and n + 1 to π when
dealing with permutations.
The number of times the substring ‘ab’ occurs in S is denoted by fab (S), where
a ∈ {α, 0, 1} and b ∈ {0, 1, ω}.
The number of reversal breakpoints (over a binary alphabet), rb(S, T ), is therefore
rb(S, T ) = |f00 (S) − f00 (T )| + |f11 (S) − f11 (T )|
+|f01 (S) + f10 (S) − f01 (T ) − f10 (T )|
+|fα0 (S) − fα0 (T )| + |fα1 (S) − fα1 (T )|
+|f0ω (S) − f0ω (T )| + |f1ω (S) − f1ω (T )|.
Clearly, if S = T , then rb(S, T ) = 0. However, it is possible to have rb(S, T ) = 0,
even when S 6= T . For example, if S = 100101 and T = 101001, then rb(S, T ) = 0.
We now derive a lower bound for reversal distance based on these breakpoints.
Lemma 5.3.1 Suppose that S 0 is obtained from S by a single reversal. Then
rb(S 0 , T ) ≥ rb(S, T ) − 4.
Proof A reversal on S cuts two substrings of length two in the extended version
of S. Suppose that the first of these substrings is ab and the other is cd. Recall
that the frequency counts of 01 substrings and 10 substrings are added together when
counting breakpoints. So the frequency counts of substrings in S 0 will be the same
as the frequency counts of substrings in S except that S 0 contains one less ab and cd
and one more ac and bd. Given the definition of rb(S 0 , T ), this means that rb(S 0 , T ) ≥
rb(S, T ) − 4. 2
This lemma can be used to easily deduce the following lower bound for reversal
distance.
Theorem 5.3.1 Let S and T be related binary strings. Then
rd(S, T ) ≥ drb(S, T )/4e.
Chapter 5. Sorting Strings by Global Transformations
118
Unfortunately this lower bound is not tight. For example, if S = 0011000111 and
T = 1110011000, then rb(S, T ) = 4, but rd(S, T ) = 2. Note also that sometimes it is
impossible to apply a reversal that will reduce the number of reversal breakpoints. For
example, rb(S, T ) = 0, but S 6= T .
5.3.2
An upper bound
In this section a simple upper bound is derived for reversal distance.
Lemma 5.3.2 Let S and T be related strings of length n, such that S 6= T . Then it is
possible either:
a) to apply a reversal on S resulting in the string S 0 , such that lcp(S 0 , T ) + lcs(S 0 , T ) ≥
lcp(S, T ) + lcs(S, T ) + 2, or
b) to apply a reversal on T resulting in the string T 0 such that lcp(S, T 0 ) + lcp(S, T 0 ) ≥
lcp(S, T ) + lcs(S, T ) + 2.
Proof Without loss of generality, it can be assumed that S(1) = 0, since otherwise
we could consider S −1 and T −1 , and apply the resulting reversal to S and T . Further,
it can be assumed that S(1) 6= T (1), and S(n) 6= T (n), because otherwise we could
reduce n by removing any common prefix or suffix from both strings.
The reversal applied to S or T depends on S and T . We describe seven cases, and
show a reversal with the required property for each one. The seven cases cover all
possibilities for S and T , though the cases are not necessarily mutually exclusive.
Case (i). S(n) = 1: then S = 0 . . . 1 and T = 1 . . . 0, so take S 0 = [0 . . . 1].
Case (ii). T (2) = 0: then S = 0 . . . 0, and T = 10 . . . 1, so take S 0 = [0 ∼ 01] . . . 0,
where the ‘01’ substring at the end of the reversal is the first ‘01’ substring in S.
Case (iii). T (n − 1) = 0: then S = 0 . . . 0, and T = 11 . . . 01, so, by considering S,
and T , this case can be dealt with in a similar way to Case (ii).
Case (iv). f11 (S) > 0: then S = 0 . . . 11 . . . 0 and T = 11 . . . 11, so take S 0 =
[0 . . . 11] . . . 00, where the ‘11’ substring at the end of the reversal is the first ‘11’
substring in S.
Case (v). S(2) = 1: then S = 01 . . . 0 and T = 11 . . . 11 so, by considering S −1 , and
−1
T , this case can be dealt with in a similar way to Case (ii).
Case (vi). S(n − 1) = 1: then S = 00 . . . 10 and T = 11 . . . 11 so, by considering
−1
−1
S , and T , this case can be dealt with in a similar way to Case (ii).
Case (vii). f00 (T ) > 0: then S = 00 . . . 00 and T = 11 . . . 00 . . . 11, so, by considering S −1 , and T −1 , this case can be dealt with in a similar way to Case (iv).
These cases are exhaustive because Cases (i) to (iv) can only fail to apply if S
contains more zeros than ones, whereas Cases (i), and (v) to (vii) can only fail to apply
if T contains more ones than zeros. 2
Theorem 5.3.2 Let S and T be related binary strings of length n. Then
rd(S, T ) ≤ bn/2c.
119
Chapter 5. Sorting Strings by Global Transformations
Proof Lemma 5.3.2 describes a way to increase the combined length of the common
prefix and suffix of S and T by at least two using a single reversal. So a sequence of
bn/2c such reversals will be enough to transform S into T . 2
For example, if S = 010101010, and T = 110000011, then applying reversals as
described in the proof of Lemma 5.3.2 results in the following sequence of strings. (In
this example the substrings to be reversed are underlined).
010101010
011001010
011000101
011000011
110000011
Note that the first reversal found as described in the proof of Lemma 5.3.2 is the one
that reverses the first three characters of T , and so it is the last reversal shown in the
illustration.
5.3.3
Reversal diameter of Bn
The reversal diameter of Bn , rD2 (n), is defined to be the maximum value of rd(S, T )
over all related binary strings S and T of length n. More formally
rD2 (n) = max{rd(S, T ) : S, T are related binary strings of length n}.
Lemma 5.3.3 For all k ≥ 1, rd(Bk , Ck ) = k, and rd(0 ++ Bk , 0 ++ Ck ) = k.
Proof This follows at once by application of Theorem 5.3.1, and Theorem 5.3.2 to
these strings. 2
Theorem 5.3.3 For all n ≥ 1, rD2 (n) = bn/2c.
Proof This is an immediate consequence of Theorem 5.3.2 and Lemma 5.3.3. 2
Theorem 5.3.4 Let S and T be related binary strings of length 2n ≥ 6.
rd(S, T ) = n, if and only if S and T are isomorphic to Cn and Bn .
Then
Proof We prove this theorem by induction. The base case is when 2n = 6. Then,
by complete search, it may be verified that rd(S, T ) = 3 if and only if S and T are
isomorphic to B3 and C3 . Now suppose that the theorem holds for n ≤ k. Let S and
T be strings of length 2k + 2 such that rd(S, T ) = k + 1. We show that S and T are
isomorphic to Ck+1 and Bk+1 .
Chapter 5. Sorting Strings by Global Transformations
120
We can assume, without loss of generality, that S(1) = 0. By Lemma 5.3.2, we can
apply a reversal to S or T that increases the combined length of the common prefix and
suffix by at least two. By the proof of Lemma 5.3.2 we can assume that the reversal
is applied to S resulting in the string S 0 . It must be that lcp(S 0 , T ) + lcs(S 0 , T ) = 2,
and rd(S 0 , T ) = k, since any alternative would contradict rd(S, T ) = k + 1. Let Se0
and Te be the strings S 0 and T excluding common prefix and suffix. By the induction
hypothesis, Se0 and Te must be isomorphic to Bk and Ck . Therefore, since Bk−1 = Bk ,
and Ck−1 = Ck , either:
a)
b)
c)
d)
Se0
Se0
Se0
Se0
= Bk
= Ck
= Bk
= Ck
and Te = Ck ; or
and Te = Bk ; or
and Te = Ck ; or
and Te = Bk .
By the proof of Lemma 5.3.2 there are essentially three ways that the reversal can
be applied to S, as typified by cases (i), (ii) and (iv) in that proof. We take each
of these three cases in turn, and show that for the four possible values of S e0 and Te ,
rd(S, T ) = k + 1 if and only if S and T are isomorphic to Bk+1 and Ck+1 .
Case (i). S = 0 . . . 1, T = 1 . . . 0. In this case the whole of S is reversed. We
show that cases a) and b) for Se0 and Te lead to contradictions, whereas cases c) and
d) establish the induction step.
a) S = 0 ++ Bk ++ 1 = 011 ∼ 100 ∼ 01, T = 1 ++ Ck ++ 0 = 11010 ∼ 100. Suppose
that instead of applying the reversal of Lemma 5.3.2 we apply the reversal [011] ∼
100 ∼ 01 to obtain S 00 . This reversal extends the common prefix by three characters,
so rd(S 00 , T ) < k. So rd(S, T ) < k + 1, a contradiction.
b) S = 0 ++ Ck ++ 1 = 00101 ∼ 011, T = 1 ++ Bk ++ 0 = 100 ∼ 011 ∼ 10. These
strings are isomorphic to the strings in a), so rd(S, T ) < k + 1, a contradiction.
c) S = 0 ++ Bk ++ 1 = Bk+1 , T = 1 ++ Ck ++ 0 = Ck+1 .
d) S = 0 ++ Ck ++ 1 = Ck+1 , T = 1 ++ Bk ++ 0 = Bk+1 .
Case (ii). S = 0 . . . 0, T = 10 . . . 1. In this case the reversal results in a string S 0
that has prefix ‘10’. So Se0 and Te must be suffixes of S 0 and T . Now since T ends with
a 1, only cases b) and c) need to be considered. Both cases lead to a contradiction.
b) S 0 = 10 ++ Ck = 1010 ∼ 10, and T = 10 ++ Bk = 1000 ∼ 011 ∼ 1. The reversal
on S ends with the first ‘01’ substring in S, so S = 011010 ∼ 10. Then the reversal
[01101]0 ∼ 10 produces string S 00 that has rd(S 00 , T ) < k by the induction hypothesis.
So rd(S, T ) < k + 1, a contradiction.
c) S 0 = 10 ++ Bk = 1011 ∼ 100 ∼ 0, and T = 10 ++ Ck = 100101 ∼ 01. Given
the nature of the reversal on S, S = 0111 ∼ 100 ∼ 0. Then the reversal 0111 ∼
[100 ∼ 0] results in a string S 00 , that has rd(S 00 , T ) < k by the induction hypothesis.
So rd(S, T ) < k + 1, a contradiction.
Case (iv). S = 0 . . . 11 . . . 0, and T = 11 . . . 11. In this case the reversal is applied
to S to obtain a string S 0 that has prefix 11. So Se0 and Te must be suffixes of S 0 and
Chapter 5. Sorting Strings by Global Transformations
121
T . Now, since T ends with ‘11’ only case b) need be considered. However, in fact, even
this case cannot occur.
b) S 0 = 11 ++ Ck = 111010 ∼ 10, and T = 11 ++ Bk = 1100 ∼ 011 ∼ 1. But then
the reversal on S could not have moved the first ‘11’ substring in S. So this case cannot
occur.
So rd(S, T ) = k + 1, if and only if S and T are isomorphic to Bk+1 and Ck+1 .
Therefore, by induction, we have proved the theorem. 2
Theorem 5.3.4 describes the strings of length n that achieve the reversal diameter,
when n is even. When n is odd, significantly more pairs of strings achieve the reversal
diameter.
5.3.4
Sorting by reversals
Let Sı denote the string that is related to S and consists only of a block of zeros,
followed by a block of ones. For example, if S = 01100110 then Sı = 00001111.
Then determining rd(S, Sı ) is an analogue of determining the reversal distance of a
permutation. We show that rd(S, Sı ) can be determined in polynomial-time.
Recall that z(S) denotes the number of blocks of zeros contained in S. Obviously,
z(Sı ) = 1 (unless S does not contain any zeros). The following lemma can be verified
easily.
Lemma 5.3.4 Let S 0 be a string obtained from S by a single reversal. Then
z(S 0 ) ≥ z(S) − 1.
With this lemma we can determine rd(S, Sı ).
Theorem 5.3.5 For any binary string S,
rd(S, Sı ) =
(
z(S) − 1, if S(1) = 0,
z(S),
otherwise.
Proof By Lemma 5.3.4, rd(S, Sı ) ≥ z(S) − 1. If S(1) = 0, then z(S) − 1 reversals
of the form 0 ∼ 0[1 . . . 0]b . . ., where b ∈ {1, ω}, transform S into Sı . If S(1) = 1,
then an extra reversal is required, because it is impossible to change the character at
the beginning to a 0, and also reduce the value of z. This bound can be achieved by
performing the reversal [1 . . . 0]b . . ., before performing the z(S) − 1 reversals described
for when S(1) = 0. 2
The distance described in Theorem 5.3.5 can be calculated easily in polynomialtime. In Section 5.7 it is shown that, in general, determining rd(S, T ) is NP-hard.
Chapter 5. Sorting Strings by Global Transformations
5.4
122
Prefix-reversal distance between binary strings
In this section, we present an upper bound and a lower bound for prefix-reversal distance between binary strings. These bounds are used to determine the prefix-reversal
diameter of Bn , and to identify some strings that achieve the prefix-reversal diameter.
A restricted version of the problem, that is in some ways analogous to the problem of
sorting by prefix-reversals, is shown to be solvable in polynomial-time.
5.4.1
A lower bound
The lower bound for prefix-reversal distance is based on blocks instead of breakpoints.
It is possible to base the lower bound on breakpoints, but such a bound is not as
elegant because the first character needs to be dealt with in a special way. Note, it
is possible to obtain a lower bound for reversal distance (and transposition distance
and block-interchange distance) based on blocks instead of breakpoints. However, in
general, such a bound is not as tight as the bound based on breakpoints.
The following lemma can be verified easily.
Lemma 5.4.1 Suppose that S 0 is obtained from S by a single prefix-reversal. Then
b(S) − b(S 0 ) ∈ {−1, 0, 1}.
This lemma can be used to deduce the following lower bound for prefix-reversal
distance.
Theorem 5.4.1 Let S and T be related binary strings of length n. Then
prd(S, T ) ≥
(
|b(S) − b(T )| + 1, if S(n) 6= T (n),
|b(S) − b(T )|,
otherwise.
Proof By Lemma 5.4.1 at least |b(S) − b(T )| prefix-reversals are required, just to
ensure that both strings contain the same number of blocks. If S(n) 6= T (n) then at
one point in the transformation from S into T , the whole string will have to be reversed.
Such a prefix-reversal cannot change the value of b, so at least one more prefix-reversal
is required in such instances. 2
This lower bound is not exact. For example, if S = 011001 and T = 101100, then
the bound states prd(S, T ) ≥ 1, whereas prd(S, T ) = 2.
5.4.2
An upper bound
In this section a simple upper bound is obtained for prefix-reversal distance.
Chapter 5. Sorting Strings by Global Transformations
123
Lemma 5.4.2 Let S and T be related binary strings, such that S 6= T . Then it is
possible either:
a) to apply a prefix-reversal to S or T , obtaining the string S 0 or T 0 , such that
lcs(S 0 , T ) ≥ lcs(S, T ) + 1, or lcs(S, T 0 ) ≥ lcs(S, T ) + 1, or
b) to apply a sequence of two prefix-reversals to S or T resulting in the string S 00 or
T 00 , such that lcs(S 00 , T ) ≥ lcs(S, T ) + 2, or lcs(S, T 00 ) ≥ lcs(S, T ) + 2.
Proof It can be assumed without loss of generality that S(1) = 0, and S(n) 6= T (n).
The prefix-reversals that are applied to S or T depend on S and T . We describe
six cases, and show prefix-reversals with the required property in each case. The six
cases cover all possibilities for S and T , though the cases are not necessarily mutually
exclusive.
Case (i). T (n) = 0: then S = 0 . . . 1 and T = . . . 0, so take S 0 = [0 . . . 1].
Case (ii). T (1) = 0: then S = 0 . . . 0, and T = 0 . . . 1, so swapping the roles of S
and T , this case is similar to Case (i).
Case (iii). T (n − 1) = 0: then S = 0 . . . 0, and T = 1 . . . 01, so take S 0 = [0 ∼
01] . . . 0, then S 00 = [10 . . . 0], where the ‘01’ substring at the end of the first prefixreversal is the first ‘01’ substring in S.
Case (iv). S(n − 1) = 1: then S = 0 . . . 10, and T = 1 . . . 11, so, by considering
−1
S , and T −1 , this case is similar to Case (iii).
Case (v). f11 (S) > 0: then S = 0 . . . 11 . . . 00, and T = 1 . . . 11, so take S 0 =
[0 . . . 11] . . . 00, then S 00 = [11 . . . 00], where the ‘11’ substring at the end of the first
prefix-reversal is the first ‘11’ substring in S.
Case (vi). f00 (T ) > 0: then S = 0 . . . 00, and T = 1 . . . 00 . . . 11, so by considering
−1
S , and T −1 , this case is similar to Case (v).
These cases are exhaustive because Cases (i) to (v) can only fail to apply if S
contains more zeros than ones, whereas Cases (i) to (iv), and (vi) can only fail to apply
if T contains more ones than zeros. 2
Theorem 5.4.2 Let S and T be related binary strings of length n. Then
prd(S, T ) ≤ n − 1.
Proof The prefix-reversals described by Lemma 5.4.2 extend the common suffix of the
two strings by at least one character on average. The sequence of two prefix-reversals
is only selected if the strings are at least three characters long. Therefore, at most
n − 1 prefix-reversals like those indicated by Lemma 5.4.2 will be required to transform
S into T . 2
5.4.3
Prefix-reversal diameter of Bn
The prefix-reversal diameter of Bn , prD2 (n), is defined to be the maximum value of
prd(S, T ) over all related binary strings S and T of length n. More formally
prD2 (n) = max{prd(S, T ) : S, T are related binary strings of length n}.
124
Chapter 5. Sorting Strings by Global Transformations
Lemma 5.4.3 For all k ≥ 1, prd(Bk , Ck ) = 2k − 1, and prd(0 ++ Bk , 0 ++ Ck ) = 2k.
Proof This follows at once by application of Theorem 5.4.1 and Theorem 5.4.2 to
these strings. 2
Theorem 5.4.3 For all n ≥ 1, prD2 (n) = n − 1.
Proof This is a simple consequence of Theorem 5.4.2 and Lemma 5.4.3. 2
Theorem 5.4.4 Let S and T be related strings of length n ≥ 2. Then prd(S, T ) =
n − 1, when n is even, if and only if S and T are prefix isomorphic to C n/2 and Bn/2 ,
and prd(S, T ) = n − 1, when n is odd, if and only if S and T are prefix isomorphic to
0 ++ Cbn/2c and 0 ++ Bbn/2c .
Proof We prove this theorem by induction. The base cases are when n = 2 and n = 3.
It is easy to verify the theorem for these values of n. Now suppose that the theorem
holds for n ≤ k. Let S and T be strings of length k + 1 such that prd(S, T ) = k. We
show that S and T are isomorphic to the strings indicated.
We can assume, without loss of generality, that S(1) = 0. By Lemma 5.4.2 we can
apply a prefix-reversal or two prefix-reversals to S or T , in such a way that each prefixreversal, on average, increases the common suffix of the two strings by a character.
By the proof of Lemma 5.4.2, we can assume that r prefix-reversals are applied to S,
resulting in the string S ∗ . It must be that lcs(S ∗ , T ) = r, and prd(S ∗ , T ) = k − r,
since any alternative would contradict the choice of S, and T . Let Se∗ , and Te be the
strings S ∗ and T excluding the common suffix. By the induction hypothesis, Se∗ and
Te must be prefix isomorphic to the strings indicated in the statement of the lemma.
Let l = k + 1 − r. Then l is the length of Se∗ and Te . If l is even then either:
a) Se∗ = B l , and Te = C l ,
2
2
2
2
b) Se∗ = C l , and Te = B l ,
c)
−1
Se∗ = B −1
l , and Te = C l , or
2
2
2
2
−1
d) Se∗ = C −1
l , and Te = B l .
If l is odd then either:
e)
Se∗ = 0 ++ B l−1 , and Te = 0 ++ C l−1 ,
f)
Se∗ = 0 ++ C l−1 , and Te = 0 ++ B l−1 ,
2
2
2
2
2
2
2
2
g) Se∗ = 1 ++ B −1
+ C −1
l−1 , and Te = 1 +
l−1 , or
h) Se∗ = 1 ++ C −1
+ B −1
l−1 , and Te = 1 +
l−1 .
125
Chapter 5. Sorting Strings by Global Transformations
By the proof of Lemma 5.4.2, there are essentially three different ways that the
prefix-reversal or prefix-reversals could be applied to S, as typified by cases (i), (iii)
and (v) in that proof. We take each case in turn, and show that S and T must be
prefix isomorphic to the strings indicated above if prd(S, T ) = n − 1.
First of all, suppose that k + 1 is even.
Case (i). S = 0 . . . 1 and T = . . . 0. In this case the prefix-reversal reverses the
whole of S. Only one prefix-reversal is applied so l is odd. Only two cases g) and h)
need to be considered since S ∗ must begin with 1. Both cases establish the induction
step.
+0 =
g) S = 0 ++ B k−1 ++ 1 = 000 ∼ 011 ∼ 11 = B k+1 and T = 1 ++ C −1
k−1 +
2
2
2
10101 ∼ 010 = C k+1 .
2
+ B −1
+ 0 = 111 ∼
h) S = 0 ++ C k−1 ++ 1 = 01010 ∼ 101 = C −1
k−1 +
k+1 and T = 1 +
2
2
100 ∼ 00 = B −1
k+1 .
2
2
Case (iii). S = 0 . . . 0, and T = 1 . . . 01. In this case a sequence of two prefixreversals is applied to S to obtain S 00 , where S 0 = [0 ∼ 01] . . . 0, and S 00 = [10 . . . 0].
Therefore l is even. Only two cases a) and d) need to be considered, since T must
begin with 1. However both cases lead to a contradiction.
a) S 00 = B k−1 ++ 01 = 00 ∼ 011 ∼ 101, and T = C k−1 ++ 01 = 1010 ∼ 1001.
2
2
So S = 0111 ∼ 100 ∼ 0. Take T 0 = [1010 . . . 100]1, then T 00 = [001 . . . 011]. Then
prd(S, T 00 ) < k − 2 by the induction hypothesis, so prd(S, T ) < k.
+ 01 = 0101 ∼ 0101, and T = B −1
+ 01 = 11 ∼ 100 ∼ 001.
d) S 00 = C −1
k−1 +
k−1 +
2
2
So S = 0110 . . . 1010. Take T 0 = [11 . . . 10]0 . . . 01, and T 00 = [011 . . . 10 . . . 01]. Then
prd(S, T 00 ) < k − 2 by the induction hypothesis, so prd(S, T ) < k.
Case (v). S = 0 . . . 11 . . . 00, and T = 1 . . . 11. In this case two prefix-reversals are
applied to S to obtain S 00 , where S 0 = [0 . . . 11] . . . 00, and S 00 = [11 . . . 00]. Therefore
l is even. Only case a) needs to be considered, since S 00 begins with ‘00’, but, in fact,
even this case cannot occur.
a) S 00 = B k−1 ++ 11 = 00 ∼ 011 ∼ 111 and T = C k−1 ++ 11 = 1010 ∼ 1011. But
2
2
the first prefix-reversal on S should end with the first ‘11’ substring in S. However,
that would mean that S 00 could not possibly have the form shown. So this case cannot
occur.
Now, let us suppose k + 1 is odd.
Case (i). S = 0 . . . 1 and T = . . . 0. The prefix-reversal reverses the whole of S.
Therefore l is even. Only case b) and c) need to be considered, since S 0 begins with 1.
Case b) leads to a contradiction, and case c) establishes the induction step.
= 00101 ∼ 01 and T = B k ++ 0 = 00 ∼ 011 ∼ 10. Take
b) S = 0 ++ C −1
k
2
2
T 0 = [00 ∼ 01]1 ∼ 10, and then T 00 = [100 ∼ 011 ∼ 10]. Now prd(S, T 00 ) < k − 2, by
the inductive hypothesis, so prd(S, T ) < k.
c) S = 0 ++ B k = 000 ∼ 011 . . . 1 and T = C −1
+ 0 = 0101 ∼ 010 = 0 ++ C k .
k +
2
2
2
Chapter 5. Sorting Strings by Global Transformations
126
Case (iii). S = 0 . . . 0, and T = 1 . . . 01. In this case two prefix-reversals are applied
to S to obtain S 00 , where S 0 = [0 . . . 01] . . . 0, and S 00 = [10 . . . 0]. Therefore l is odd.
But T (1) = 1, and S 00 (1) = 0 and so by the induction hypothesis prd(S 00 , T ) < k − 2.
Therefore, prd(S, T ) < k.
Case (v). S = 0 . . . 11 . . . 00, and T = 1 . . . 11. In this case two prefix-reversals are
applied to S to obtain S 00 , where S 0 = [0 . . . 11] . . . 00 and S 00 = [11 . . . 00]. Therefore l
is odd. But T (1) = 1, and S 00 (1) = 0 and so by the induction hypothesis prd(S 00 , T ) <
k − 2. Therefore, prd(S, T ) < k.
So, by induction, we have proved the theorem. 2
5.4.4
Sorting by prefix-reversals
We show that prd(S, Sı ) can be determined in polynomial-time.
The following lemma can be verified easily.
Lemma 5.4.4 Let S 0 be a string obtained from S by a single prefix-reversal. Then
z(S 0 ) ≥
(
z(S) − 1, if S(1) = 0,
z(S),
otherwise.
With this lemma we can determine prd(S, Sı ).
Theorem 5.4.5 For any binary string S,
prd(S, Sı ) =
(
2(z(S) − 1), if S(1) = 0,
2z(S) − 1,
otherwise.
Proof If a prefix-reversal reduces the value of z, then it must bring 1 to the front of
the string, and then by Lemma 5.4.4 the next prefix-reversal cannot reduce the value of
z. Note also that the final prefix-reversal cannot reduce the value of z because it must
bring 0 to the front. Therefore prd(S, Sı ) ≥ 2(z(S) − 1). If S(1) = 0, then z(S) − 1
prefix-reversals of the form [0 ∼ 01 ∼ 1]0 . . . interleaved with z(S)−1 prefix-reversals of
the form [1 ∼ 10 ∼ 0]b . . ., where b ∈ {1, ω}, transform S into Sı . If S(1) = 1, then an
extra prefix-reversal is required, since the first prefix-reversal cannot reduce the value
of z. This bound can be achieved by performing the prefix-reversal [1 ∼ 10 ∼ 0]b . . .,
where b ∈ {1, ω}, before performing the 2(z(S) − 1) prefix-reversals just described for
when S(1) = 0. 2
The distance described in Theorem 5.4.5 can be calculated easily in polynomialtime. The question of whether, in general, the prefix-reversal distance between any
two strings can be calculated in polynomial-time remains open.
Chapter 5. Sorting Strings by Global Transformations
5.5
127
Transposition distance between binary strings
In this section, we present an upper bound and a lower bound for transposition distance
between binary strings. These bounds are used to determine the transposition diameter
of Bn , and identify some strings that achieve the diameter. A restricted version of the
problem, that is in some ways analogous to the problem of sorting by transpositions,
is shown to be solvable in polynomial-time.
5.5.1
A lower bound
In this section, breakpoints are used to obtain a lower bound for transposition distance.
Transposition breakpoints are defined in a similar way to reversal breakpoints. For
example, if S contains more 11 substrings than T then each extra 11 substring contributes a breakpoint. However a crucial difference between reversal breakpoints and
transposition breakpoints is that 01 substrings and 10 substrings are counted separately. As before prepend α and append ω to each string.
The number of transposition breakpoints (over a binary alphabet) is therefore
tb(S, T ) = |f00 (S) − f00 (T )| + |f01 (S) − f01 (T )| + |f10 (S) − f10 (T )|
+|f11 (S) − f11 (T )| + |fα0 (S) − fα0 (T )| + |fα1 (S) − fα1 (T )|
+|f0ω (S) − f0ω (T )| + |f1ω (S) − f1ω (T )|.
Clearly, tb(T, T ) = 0. However it is possible that tb(S, T ) = 0, even when S 6= T .
For example, if S = 101001 and T = 100101, then tb(S, T ) = 0.
Lemma 5.5.1 Suppose that S 0 is obtained from S by a single transposition. Then
0
tb(S , T ) ≥
(
tb(S, T ) − 6, if the transposition moves the first and last letters of S,
tb(S, T ) − 4, otherwise.
Proof The transposition must have the form . . . a[b . . . c][d . . . e]f . . ., where a ∈
{α, 0, 1}, b, c, d, e ∈ {0, 1}, and f ∈ {0, 1, ω}. The transposition results in the string
. . . a[d . . . e][b . . . c]f . . .. Now let us suppose that the none of the substrings ‘ab’, ‘cd’,
or ‘ef ’ is the same as any of the substrings ‘ad’, ‘eb’, or ‘cf ’. Then ‘cd’6=‘cf ’, so d 6= f .
Similarly, c 6= a, b 6= f , and e 6= a. Now suppose that a 6= α. Then c = e. But then
‘ef ’=‘cf ’, a contradiction. Similarly if f 6= ω then ‘ab’=‘ad’. Therefore, if a 6= α, or
f 6= ω, then at least one of the substrings ‘ab’, ‘cd’, or ‘ef ’ is the same as one of the
substrings ‘ad’, ‘eb’, or ‘cf ’.
If a transposition does move the first and last characters of S then at most three
substrings of length two may change as a result of the transposition. Therefore, in such
cases, the frequency counts of substrings in S 0 are the same as the frequency counts of
substrings in S except that the frequency counts of up to three substrings are reduced
Chapter 5. Sorting Strings by Global Transformations
128
by one, and the frequency counts of up to three substrings are increased by one. Given
the definition of tb(S 0 , T ), this means that tb(S 0 , T ) ≥ tb(S, T ) − 6.
However, if a transposition does not move the first and last characters of S, then
at most two substrings of length two may change as a result of the transposition. In
such cases tb(S 0 , T ) ≥ tb(S, T ) − 4. 2
Theorem 5.5.1 Let S and T be related binary strings, of length n. Then
td(S, T ) ≥
(
dtb(S, T )/4e,
if S(1) = T (1), or S(n) = T (n),
d(tb(S, T ) − 2)/4e, otherwise.
Proof A sequence of transpositions that transforms S into T can contain at most one
transposition that reduces the number of breakpoints by six. Such a transposition is
only possible if S(1) 6= T (1), and S(n) 6= T (n). Every other transposition can reduce
the number of breakpoints by at most four. The theorem follows easily from these
observations. 2
This lower bound is not exact. For example, if S = 011100110001 and T =
100011001110, then the bound is 1, but td(S, T ) = 2.
5.5.2
An upper bound
In this section a simple upper bound is derived for transposition distance.
Lemma 5.5.2 Let S and T be related strings of length n, such that S 6= T . Then it is
possible either:
a) to apply a transposition to S resulting in a string S 0 such that lcp(S 0 , T )+lcs(S 0 , T ) ≥
lcp(S, T ) + lcs(S, T ) + 2, or
b) to apply a transposition to T resulting in a string T 0 such that lcp(S, T 0 )+lcs(S, T 0 ) ≥
lcp(S, T ) + lcs(S, T ) + 2.
Proof Without loss of generality, it can be assumed that S(1) = 0, S(1) 6= T (1),
and S(n) 6= T (n). The transposition that is applied to S or T depends on S and T .
Three cases are described, and for each case a transposition is shown with the required
property. The three cases cover all possibilities for S and T , though the cases are not
necessarily mutually exclusive.
Case (i). S(n) = 1: Then S = 0 . . . 1 and T = 1 . . . 0, so take S 0 = [0 ∼ 0][1 . . . 1],
where the ‘01’ substring that is split apart by the transposition is the first ‘01’ substring
in S.
Case (ii). f11 (S) > 0: Then S = 0 . . . 11 . . . 0, and T = 1 . . . 1 so take S 0 = [00 ∼
01][1 . . . 0], where the ‘11’ substring that is split apart by the transposition, is the first
‘11’ substring in S.
Case (iii). f00 (T ) > 0: Then S = 0 . . . 0, and T = 1 . . . 00 . . . 1 so, by considering
−1
S , and T −1 , this case is similar to Case (ii).
129
Chapter 5. Sorting Strings by Global Transformations
These cases are sufficient to prove the lemma because Cases (i) and (ii) can only
fail to apply if S contains more zeros than ones, whereas Cases (i) and (iii) can only
fail to apply when T contains more ones than zeros. 2
Theorem 5.5.2 Let S and T be related binary strings, of length n. Then
td(S, T ) ≤ bn/2c.
Proof Lemma 5.5.2 describes a way to increase the the combined length of the common
prefix and suffix of S and T by at least two using a single transposition. So a sequence
of bn/2c such transpositions will be enough to transform S into T . 2
5.5.3
Transposition diameter of Bn
The transposition diameter of Bn , tD2 (n), is the maximum value of td(S, T ) taken over
all related binary strings of length n. More formally
tD2 (n) = max{td(S, T ) : S, T are related binary strings of length n}.
Lemma 5.5.3 For all k ≥ 1, td(Bk , Ck ) = k, and td(0 ++ Bk , 0 ++ Ck ) = k.
Proof In both cases, this follows at once by application of Theorem 5.5.1, and Theorem
5.5.2. 2
Theorem 5.5.3 For all n ≥ 1, tD2 (n) = bn/2c.
Proof This is an immediate consequence of Theorem 5.5.2 and Lemma 5.5.3. 2
Theorem 5.5.4 Let S and T be related binary strings of length 2n ≥ 4.
td(S, T ) = n, if and only if S and T are isomorphic to Cn and Bn .
Then
Proof We prove this theorem by induction. The base case is when 2n = 4. Then
the only strings for which td(S, T ) = 2 are isomorphic to B2 and C2 . Now suppose
that the theorem holds for n ≤ k. Let S and T be strings of length 2k + 2, such that
td(S, T ) = k + 1. We show that S and T are isomorphic to Ck+1 and Bk+1 .
We can assume, without loss of generality, that S(1) = 0. By the proof of Lemma
5.5.2 we can assume that the transposition is applied to S resulting in the string S 0 . It
must be that lcp(S 0 , T ) + lcs(S 0 , T ) = 2, and td(S 0 , T ) = k, since any alternative would
contradict td(S, T ) = k + 1. Let Se0 and Te be the strings S 0 and T excluding common
prefix and suffix. By the induction hypothesis, Se0 and Te must be isomorphic to Bk
and Ck . Therefore, either:
Chapter 5. Sorting Strings by Global Transformations
a)
b)
c)
d)
Se0
Se0
Se0
Se0
= Bk
= Ck
= Bk
= Ck
130
and Te = Ck ; or
and Te = Bk ; or
and Te = Ck ; or
and Te = Bk .
By the proof of Lemma 5.5.2, there are essentially two ways that the transposition
can be applied to S, as typified by cases (i) and (ii) in that proof. We take each case
in turn, and show that for the four possible values of Se0 and Te , td(S, T ) = k + 1 if and
only if S and T are isomorphic to Bk+1 and Ck+1 .
Case (i). S = 0 . . . 1, T = 1 . . . 0. In this case the transposition moves the block of
zeros at the front of S to the end. We show that cases c) and d) for S e0 and Te establish
the induction step, whereas cases a) and b) lead to contradictions.
a) S 0 = 1 ++ Bk ++ 0 = 100 ∼ 011 ∼ 10 and T = 1 ++ Ck ++ 0 = 11010 ∼ 100. So
S = 0100 ∼ 011 ∼ 1. But then the transposition 0[100][. . . 011. . . 1] produces a string
S 00 such that td(S 00 , T ) < k, by the induction hypothesis. So td(S, T ) < k + 1, giving a
contradiction.
b) S 0 = 1 ++ Ck ++ 0 = 11010 ∼ 100 and T = 1 ++ Bk ++ 0 = 100 ∼ 011 ∼ 10. Then
S = 0011010 . . . 01. But then the transposition [0011010 . . . 0][1] produces a string S 00
such that td(S 00 , T ) < k, by the induction hypothesis. So td(S, T ) < k + 1, giving a
contradiction.
c) S 0 = 1 ++ Bk ++ 0 = Bk+1 and T = 1 ++ Ck ++ 0 = Ck+1 . So S = Bk+1 .
d) S 0 = 1 ++ Ck ++ 0 = Ck+1 and T = 1 ++ Bk ++ 0 = Bk+1 . So S = Ck+1 .
Case (ii). S = 0 . . . 11 . . . 0, T = 1 . . . 1. In this case the transposition is applied to
S to obtain S 0 = [0 . . . 1][1 . . . 0]. Now S 0 must contain 00, so only cases a) and c) need
to be considered. However, both cases lead to contradictions.
a) S 0 = 1 ++ Bk ++ 1 = 100 ∼ 011 ∼ 11 and T = 1 ++ Ck ++ 1 = 11010 ∼ 101.
However the transposition that splits the first ‘11’ in S cannot produce a string like
S 0 . So this case cannot occur.
c) S 0 = 1 ++ Bk ++ 1 = 111 ∼ 100 ∼ 01 and T = 1 ++ Ck ++ 1 = 10101 ∼ 011. Then
S = 00 ∼ 011 ∼ 100 ∼ 0. However the transposition 00 ∼ 011 ∼ [11][00 ∼ 0] produces
a string S 00 such that td(S 00 , T ) < k, by the induction hypothesis. So td(S, T ) < k + 1,
giving a contradiction.
So td(S, T ) = k + 1 if and only if S and T are isomorphic to Bk+1 and Ck+1 .
Therefore by induction the theorem is true. 2
5.5.4
Sorting by transpositions
We show that td(S, Sı ) can be determined in polynomial-time. The following lemma
can be verified easily.
Lemma 5.5.4 Let S 0 be a string obtained from S by a single transposition. Then
z(S 0 ) ≥ z(S) − 1.
Chapter 5. Sorting Strings by Global Transformations
131
With this lemma we can determine td(S, Sı ).
Theorem 5.5.5 For any binary string S,
td(S, Sı ) =
(
z(S) − 1, if S(1) = 0,
z(S),
otherwise.
Proof By Lemma 5.5.4, td(S, Sı ) ≥ z(S) − 1. If S(1) = 0, then z(S) − 1 transpositions
of the form 0 ∼ 0[1 . . . 1][0 ∼ 0]b . . ., where b ∈ {1, ω}, transform S into Sı . If S(1) = 1,
an extra transposition is required, because is is impossible to change the character at
the front of the string to 0 with a transposition, and also reduce the value of z. The
bound in this case can be achieved by performing the transposition [1 . . . 1][0 ∼ 0]b . . .,
where b ∈ {1, ω}, before the sequence of transpositions used when S(1) = 0. 2
The distance described in Theorem 5.5.5 can be calculated easily in polynomialtime. The question of whether, in general, the transposition distance between any two
strings can be calculated in polynomial-time remains open.
5.6
Block-Interchange distance between binary strings
In this section, we present a lower bound and an upper bound for block-interchange
distance between binary strings. These bounds are then used to determine the blockinterchange diameter, and also to identify some strings that achieve the diameter. A
restricted version of the problem, that is in some ways analogous to sorting permutations by block-interchanges, is shown to be solvable in polynomial-time. However, in
Section 5.7 the general problem of determining block-interchange distance between two
strings is shown to be NP-hard.
5.6.1
A lower bound
In this section, the concept of a breakpoint is adapted to obtain a lower bound for
block-interchange distance.
Block-interchange breakpoints are defined in exactly the same way as transposition
breakpoints. So the number of block-interchange breakpoints (over a binary alphabet)
is
bb(S, T ) = |f00 (S) − f00 (T )| + |f01 (S) − f01 (T )| + |f10 (S) − f10 (T )|
+|f11 (S) − f11 (T )| + |fα0 (S) − fα0 (T )| + |fα1 (S) − fα1 (T )|
+|f0ω (S) − f0ω (T )| + |f1ω (S) − f1ω (T )|.
Clearly, bb(T, T ) = 0. However it is possible that bb(S, T ) = 0, even when S 6= T .
For example, if S = 101001 and T = 100101, then bb(S, T ) = 0.
Chapter 5. Sorting Strings by Global Transformations
132
Lemma 5.6.1 Suppose that S 0 is obtained from S by a single block-interchange. Then
bb(S 0 , T ) ≥ bb(S, T ) − 8.
Proof Four substrings of length two may change as the result of a block-interchange.
So the frequency counts of substrings in S 0 are the same as the frequency counts of
substrings in S except that the frequency counts of the four removed substrings are
reduced by one, and the frequency counts of the four new substrings are increased by
one. Given the definition of bb(S 0 , T ), this means that bb(S 0 , T ) ≥ bb(S, T ) − 8. 2
The following lower bound for block-interchange distance can be deduced easily
from the above lemma.
Theorem 5.6.1 Let S and T be related binary strings. Then
bd(S, T ) ≥ dbb(S, T )/8e.
This lower bound is not exact. For example, if S = 011100110001 and T =
100011001110, then the bound is one , whereas bd(S, T ) = 2. In fact this lower bound
can be strengthened, since a block interchange can only reduce the number of breakpoints by eight if f00 and f11 are both reduced by two, and f01 and f10 are both
increased by two, or vice versa. However, this lower bound is sufficiently tight for our
purposes.
5.6.2
An upper bound
In this section, a simple upper bound is derived for block-interchange distance.
Lemma 5.6.2 Let S and T be related binary strings of length n, such that S 6= T and
n ≥ 4. Then it is possible to apply either:
a) a block-interchange to S obtaining S 0 such that lcp(S 0 , T ) + lcs(S 0 , T ) ≥ lcp(S, T ) +
lcs(S, T ) + 4, or
b) a block-interchange to T obtaining T 0 such that lcp(S, T 0 ) + lcs(S, T 0 ) ≥ lcp(S, T ) +
lcs(S, T ) + 4.
Proof Without loss of generality, it can be assumed that S(1) = 0, S(1) 6= T (1),
and S(n) 6= T (n). If the strings are represented by their first two and last two characters then, up to isomorphism, all possible remaining combinations of S and T are
represented by the 18 pairs shown in Table 5.1.
If S(2) = 0 then cases (i) to (xvi) cover all the possibilities for S and T . If S ends
with 00 or 11, or T begins or ends with 00 or 11 then strings isomorphic to S and T are
covered by one of the first sixteen cases. Otherwise there are only two possible cases
remaining and these are cases (xvii) and (xviii).
Chapter 5. Sorting Strings by Global Transformations
Case
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
(xiii)
(xiv)
(xv)
(xvi)
(xvii)
(xviii)
S
00. . . 00
00. . . 00
00. . . 00
00. . . 00
00. . . 01
00. . . 01
00. . . 01
00. . . 01
00. . . 10
00. . . 10
00. . . 10
00. . . 10
00. . . 11
00. . . 11
00. . . 11
00. . . 11
01. . . 01
01. . . 10
T
10. . . 01
10. . . 11
11. . . 01
11. . . 11
10. . . 00
10. . . 10
11. . . 00
11. . . 10
10. . . 01
10. . . 11
11. . . 01
11. . . 11
10 . . . 00
10. . . 10
11. . . 00
11. . . 10
10. . . 10
10. . . 01
Table 5.1: The 18 possible combinations of S and T .
133
Chapter 5. Sorting Strings by Global Transformations
134
These eighteen cases are now investigated in turn and for each case, either a blockinterchange is demonstrated that satisfies the statement of the lemma, or the case is
shown to be similar to a previous case.
Case (i). S must contain at least two ones, so take S 0 = [00 . . . 01] . . . [10 . . . 00].
Case (ii). This case can be subdivided into two cases:
a) S contains 11 before 10: take S 0 = [00 . . . 11] . . . [10 . . . 00]; else
b) T contains 00 before 00: take T 0 = [10 . . . 00] . . . [00 . . . 11].
These cases are exhaustive, because if a) fails then S contains more zeros than ones,
whereas if b) fails then T contains at least as many ones as zeros.
Case (iii). S = 00 . . . 00 and T = 10 . . . 11 so this case is identical to Case (ii) by
symmetry.
Case (iv). This case can be subdivided into two cases:
a) S contains 11 before 11: take S 0 = [00 . . . 11] . . . [11 . . . 00]; else
b) T contains 00 before 00: take T 0 = [11 . . . 00] . . . [00 . . . 11].
These cases are exhaustive, because if a) fails then S contains more zeros than ones,
whereas if b) fails then T contains more ones than zeros.
Case (v). This case can be subdivided into two cases:
a) S ends with 001: take S 0 = [00 . . . 00][1]; else
b) S ends with 101: take S 0 = [00] . . . [101].
These cases are obviously exhaustive.
Case (vi). This case can be subdivided into two cases:
a) S contains 100: take S 0 = [00 . . . 10]0 . . . 0[1]; else
b) S ends with 101: take S 0 = [00 . . . 10][1].
These cases are exhaustive because if S does not contain 100 then S must end with
101.
Case (vii). Clearly S must contain another one since T contains two ones. So take
S 0 = [00 . . . 00]1 . . . 0[1].
Case (viii). This case can be subdivided into seven cases:
a) S contains 10 before 11: take S 0 = [00 . . . 10] . . . [11 . . . 01]; else
b) S contains 111: take S 0 = [0]0 . . . 1[11 . . . 01]; else
c) S contains 101, but not only as a suffix: take S 0 = [00 . . . 10]1 . . . 0[1]; else
d) S = 001101 and T = 110010: take S 0 = [00][11]01; else
e) T contains 01 before 00: take T 0 = [11 . . . 01] . . . [00 . . . 10]; else
f) T contains 000: take T 0 = [1]1 . . . 0[00 . . . 10]; else
g) T contains 010, but not only as a suffix: take T 0 = [11 . . . 01]0 . . . 1[0].
These cases are exhaustive, because if a), b), c) and d) fail then S contains more zeros
than ones, whereas if d), e), f) and g) fail then T contains more ones than zeros.
Case (ix). Clearly S must contain another one since T contains two ones. So take
S 0 = [00 . . . 01] . . . [10].
Case (x). This case can be subdivided into two cases:
a) S contains 11 before 10: S 0 = [00 . . . 11] . . . [10 . . . 10]; else
Chapter 5. Sorting Strings by Global Transformations
135
b) T contains 10 before 00: T 0 = [10 . . . 10] . . . [00 . . . 11].
These cases are exhaustive, because if a) fails S then contains more zeros than ones,
whereas if b) fails then T contains more ones than zeros.
Case (xi). This case can be subdivided into two cases:
a) S contains 01 before 11: S 0 = [00 . . . 01] . . . [11 . . . 10]; else
b) T contains 10 before 00: T 0 = [11 . . . 10] . . . [00 . . . 01].
These cases are exhaustive, because if a) fails S then contains more zeros than ones,
whereas if b) fails then T contains more ones than zeros.
−1
−1
Case (xii). S = 10 . . . 11 and T
= 00 . . . 00 so this case is isomorphic to Case
(ii).
Case (xiii). S = 11 . . . 00 and T = 00 . . . 01 so this case is isomorphic to Case (vii).
Case (xiv). Take S 0 = [0]0 . . . 1[1].
Case (xv). Take S 0 = [00] . . . [11].
Case (xvi). S −1 = 11 . . . 00 and T −1 = 00 . . . 01 so this case is isomorphic to Case
(vii).
Case (xvii). This case can be subdivided into two cases:
a) S contains 100: take S 0 = [01 . . . 10]0 . . . 0[1]; else
b) S ends with 101: take S 0 = [01 . . . 10][1].
These cases are exhaustive, because if S does not contain 100 then it must end with
101.
Case (xviii). Take S 0 = [01] . . . [10]. 2
Theorem 5.6.2 Let S and T be related binary strings of length n. Then
bd(S, T ) ≤ d(n − 1)/4e.
Proof Lemma 5.6.2 describes a way to extend the combined length of the common
prefix and suffix of S and T by at least four using a single block-interchange. So a
sequence of d(n − 1)/4e such block-interchanges will be enough to transform S into T .
2
5.6.3
Block-interchange diameter of Bn
The block-interchange diameter of Bn , bD2 (n), is the maximum value of bd(S, T ) taken
over all related binary strings S and T of length n. More formally
bD2 (n) = max{bd(S, T ) : S, T are related binary strings of length n}.
Lemma 5.6.3 For all k ≥ 1, bd(B2k , C2k ) = k, bd(0 ++ B2k , 0 ++ C2k ) = k, bd(B2k+1 ,
C2k+1 ) = k + 1, and bd(0 ++ B2k+1 , 0 ++ C2k+1 ) = k + 1.
Chapter 5. Sorting Strings by Global Transformations
136
Proof In each case, this follows at once by application of Theorem 5.6.1 and Theorem
5.6.2. 2
Theorem 5.6.3 For all n ≥ 1, bD2 (n) = d(n − 1)/4e.
Proof This is trivially true for n = 1. For n ≥ 2, it is an immediate consequence of
Theorem 5.6.2 and Lemma 5.6.3. 2
Conjecture 5.6.1 Let S and T be related binary strings of length 4k + 2. Then
bd(S, T ) = k + 1, if and only if S and T are isomorphic to C2k+1 and B2k+1 .
It is conjectured that for other lengths of strings, more pairs of strings achieve the
block-interchange diameter.
5.6.4
Sorting by block-interchanges
We show that bd(b, Sı ) can be determined in polynomial-time. The following lemma
can be verified easily.
Lemma 5.6.4 Let S 0 be a string obtained from S by a single block-interchange. Then
z(S 0 ) ≥ z(S) − 2.
With this lemma we can determine bd(S, Sı ).
Theorem 5.6.4 For any binary string S,
bd(S, Sı ) =
(
d(z(S) − 1)/2e, if S(1) = 0
dz(S)/2e,
otherwise
Proof By Lemma 5.6.4, bd(S, Sı ) ≥ d(z(S) − 1)/2e. If S(1) = 0, and z(S) is odd,
then d(z(S) − 1)/2e block-interchanges of the form 0 ∼ 0[1 . . . 1]0 ∼ 01 . . . 1[0 ∼ 0]b . . .,
where b ∈ {1, ω}, transform S into Sı . If S(1) = 0, and z(S) is even, then not every
block-interchange can reduce z by two, therefore at least dz(S)/2e block-interchanges
are required. This bound can be achieved with a block-interchange of the form 0 ∼
0[1 ∼ 1][0 ∼ 0]b . . ., where b ∈ {1, ω} followed by b(z(S) − 1)/2c block-interchanges like
those used above when z(S) was odd.
If S(1) = 1, an extra block-interchange may be required, because is is impossible to
change the character at the front of the string to 0 with a block-interchange, and also
reduce the value of z by two. If we apply the block-interchange [1 . . . 1]0 ∼ 01 . . . 1[0 ∼
0]b . . . (or [1 ∼ 1][0 ∼ 0]b . . ., when z(S) = 1) , where b ∈ {1, ω}, before the sequence of
d(z(S) − 2)/2e block-interchanges used when S(1) = 0, the bound in this case, can be
achieved. 2
The distance described in Theorem 5.6.4 can be calculated easily in polynomialtime. In Section 5.7 it is shown that, in general, determining bd(S, T ) is NP-hard.
Chapter 5. Sorting Strings by Global Transformations
5.7
137
NP-completeness of reversal distance and block-interchange distance
In this section, we prove that the general problem of finding the reversal distance between two strings is NP-hard, even when the strings are drawn from a binary alphabet.
We also prove a similar result for block-interchange distance.
We begin with a definition of the reversal distance problem as a decision problem
(RD):
RD
Instance: Related strings S and T of length n, over an alphabet of size
t ≥ 2, and a bound d ∈ Z + .
Question: Is rd(S, T ) ≤ d?
We transform an NP-complete problem into RD to obtain the NP-completeness
result. The problem we start from is 3-Partition:
3-Partition
Instance: A set A of 3m elements, a bound B ∈ Z + , and a size s(a) ∈ Z + for
P
each a ∈ A such that B/4 < s(a) < B/2 and such that a∈A s(a) = mB.
Question: Can A be partitioned into m disjoint sets A1 , A2 , . . . , Am such
P
that, for 1 ≤ i ≤ m, a∈Ai s(a) = B? (Note that each Ai must contain
exactly three elements from A).
However, the transformation from 3-Partition to RD is a little unusual in that it is a
pseudo-polynomial transformation. We now define pseudo-polynomial transformations,
and also present the definition of NP-complete in the strong sense.
Let I be an instance of a decision problem that involves numbers. Then Length(I)
is defined to be the size of the problem instance, and Max(I) is defined to be the
magnitude of the largest number in the problem instance. A decision problem Π is
NP-complete in the strong sense if there is a polynomial p such that Πp is NP-complete,
where Πp is the problem Π restricted to instances I with Max(I) ≤ p(Length(I)).
The following lemma is due to Garey and Johnson [GJ75] (see also Chapter 4.2 of
[GJ79]).
Lemma 5.7.1 3-Partition is NP-complete in the strong sense.
Let Π and Π0 denote arbitrary decision problems with instance sets DΠ and DΠ0 ,
and yes sets YΠ and YΠ0 , with specified functions Max and Length, and Max0 and
Length0 respectively. A pseudo-polynomial transformation from Π to Π0 is a function
f : DΠ → DΠ0 such that:
Chapter 5. Sorting Strings by Global Transformations
138
a) for all I ∈ DΠ , I ∈ YΠ if and only if f (I) ∈ YΠ0 .
b) f can be computed in time polynomial in the two variables Max(I) and
Length(I).
c) there exists a polynomial q1 such that for all I ∈ DΠ
q1 (Length0 (f (I))) ≥ Length(I).
d) there exists a two variable polynomial q2 such that, for all I ∈ DΠ
Max0 (f (I)) ≤ q2 (Max(I), Length(I)).
Lemma 5.7.2 If Π is NP-complete in the strong sense, Π0 ∈ N P , and there exists a
pseudo-polynomial transformation from Π to Π0 , then Π0 is NP-complete (in the strong
sense).
Proof This is proved in Chapter 4.2 of [GJ79]. Essentially the pseudo-polynomial
transformation is a polynomial-time transformation for instances of Π p . 2
We can now prove the NP-completeness result for reversals.
Theorem 5.7.1 RD is NP-complete, even when t = 2.
Proof RD is in NP because, given a sequence of reversals, it can easily be checked
in polynomial-time that the sequence transforms S into T and has length at most d.
By Lemma 5.7.2, the proof will be completed by demonstrating a pseudo-polynomial
transformation from 3-Partition to RD. We now describe the transformation.
Let I be an instance of 3-Partition. From this instance construct an instance, I 0 , of
3m−1 s(ai ) 3
B
8m−3 , and d = 3m − 1.
RD with S = (++i=1
0
1 ) ++ 0s(a3m ) , T = (++m
i=1 0 1) ++ 1
So the blocks of zeros in S represent elements of A, and the lengths of the blocks
represent the sizes of the elements.
We first show that rd(S, T ) ≥ 3m − 1.
Let U be a string that is related to S and T . Recall that z(U ) is the number of
blocks of zeros in the string U . Define o(U ) as the number of blocks of ones of length
one in U . Then the function f (U ) = z(U )−o(U )−1 can be viewed as a kind of distance
function between strings U and T , since f (T ) = 0. Furthermore, f (S) = 3m − 1. We
show that, if U 0 is obtained from U by applying a single reversal, then f (U 0 ) ≥ f (U )−1.
Suppose that ρ is a reversal that transforms U into U 0 with f (U 0 ) < f (U ). Then ρ
must reduce the number of blocks of zeros or increase the number of blocks of ones of
length one.
If ρ reduces the number blocks of zeros then it must have the form . . . 1[0 . . . 1]0 . . .
or . . . 0[1 . . . 0]1 . . ., so f (U )−f (U 0 ) = 1. So it is impossible for ρ to increase the number
of blocks of ones of length one as well as reduce the number of blocks of zeros.
Chapter 5. Sorting Strings by Global Transformations
139
If ρ increases the number of blocks of ones of length one then it must be of
the form . . . 01[10 . . . 0]0 . . ., or . . . 1[10 . . . 11]0 . . ., or . . . 01[1 . . . 0]11 . . ., or the mirror image of one of these three reversals. (Note, the reversals . . . 01[11 . . . 0]0 . . ., and
. . . 11[10 . . . 0]0 . . . do not reduce the value of f .) In each case f (U ) − f (U 0 ) = 1. So
rd(S, T ) ≥ 3m − 1.
Note that the first of the reversals in the previous paragraph is special because it
increases the number of blocks of length one by two, but also increases the number of
blocks of zeros. We call this kind of reversal a bad reversal.
We now show that the given transformation from 3-Partition to an instance of RD
is a pseudo-polynomial transformation. We verify the four properties, a), b), c) and d)
in turn.
For property a), we have to show that I is a yes instance of 3-Partition if and only
if rd(S, T ) ≤ 3m − 1.
We have already shown that rd(S, T ) ≥ 3m − 1. We now show that, if rd(S, T ) =
3m − 1, then no minimal length sequence of reversals that transform S into T contains
a bad reversal.
Suppose that rd(S, T ) = 3m − 1. Every reversal in a minimal length sequence
that transforms S into T must reduce the value of f by one. For a reversal to be bad
the string must contain 0110 as a substring. However S contains no such substring,
and no reversal that reduces the value of f by one can create such a substring. So if
rd(S, T ) = 3m − 1, then no minimal length sequence of reversals that transforms S
into T contains a bad reversal.
This means that, if rd(S, T ) = 3m − 1, each block of zeros in T is constructed
from three blocks of zeros in S. It follows that I is a yes instance of 3-Partition if
rd(S, T ) = 3m − 1.
Now we show that if I is a yes instance of 3-Partition then rd(S, T ) ≤ 3m − 1.
Since I is a yes instance, we can partition A into m disjoint sets A1 , . . ., Am , each
of which contains three elements, and sums to B. For each subset Ai in turn, we
can use two reversals, of the form . . . 0[1 . . . 0]a . . ., where a ∈ {1, ω} (where ω is the
special character used to denote the end of the string), to merge the three blocks of
zeros representing the elements of Ai into a single block of zeros of length B, without
affecting any other block of zeros. (Note that the reversals shown do not move the
block of zeros at the front of the string). Then we can use m − 1 reversals, of the form
. . . 01[11 . . . 0]1 . . ., to create blocks of ones of length one separating the blocks of zeros.
This sequence of reversals has length 3m − 1, so rd(S, T ) ≤ 3m − 1. This establishes
the required property a).
To prove properties b), c) and d) we need Length and Max functions for 3-Partition
P
and RD. For 3-Partition, reasonable definitions are Length(I) = |A| + a∈A dlog2 s(a)e,
and Max(I) = max{s(a) : a ∈ A} [GJ79]. For RD, reasonable definitions are, for
example, Length0 (I 0 ) = 2n + dlog2 de, and Max0 (I 0 ) = 1 (since RD is not a number
P
problem). Note that n = a∈A s(a) + |A| − 3.
Chapter 5. Sorting Strings by Global Transformations
140
Given these functions, properties b), c) and d) can be proved quite easily.
So the transformation is a pseudo-polynomial time transformation, and therefore,
by Lemma 5.7.2, RD is NP-complete. 2
The transformation just described was obtained after several other simpler transformations had been shown to fail. For example, we tried the following transformation from sorting by reversals to RD. Given a permutation π, define strings
π(i) 1) ++ 0π(n) , and T = (++n−1 0i 1) ++ 0n . One might conjecture that
S = (++n−1
i=1 0
i=1
determining the value of rd(S, T ) must also determine the value of rd(π). However,
if π = 3142, then rd(π) = 3, but S = 0001010000100, T = 0100100010000 and
rd(S, T ) = 2. So this transformation does not work.
We now prove that the general problem of finding the block-interchange distance
between two strings is NP-hard, even when the strings are drawn from a binary alphabet. The block-interchange distance problem can be defined as a decision problem
(BD), as follows:
BD
Instance: Related strings S and T of length n, over an alphabet of size
t ≥ 2, and a bound d ∈ Z + .
Question: Is bd(S, T ) ≤ d?
This time the transformation starts from a special case of 3-Partition called Odd3-Partition.
Odd-3-Partition
Instance: A set A of 3m elements, where m is odd , a bound B ∈ Z + , and
P
a size s(a) ∈ Z + for each a ∈ A with B/4 < s(a) < B/2 and a∈A s(a) =
mB.
Question: Can A be partitioned into m disjoint sets A1 , A2 , . . . , Am such
P
that, for 1 ≤ i ≤ m, a∈Ai s(a) = B? (Note that each Ai must contain
exactly three elements from A)
We now show that this special case of 3-Partition is NP-complete in the strong
sense.
Lemma 5.7.3 Odd-3-Partition is NP-complete in the strong sense.
Proof Let I be an instance of 3-Partition. We transform I into an instance of Odd3-Partition. If m is odd then I is already an instance of Odd-3-Partition. Otherwise,
let m0 = m + 1, A0 = A ∪ {a3m+1 , a3m+2 , a3m+3 }, s0 (a) = 3.s(a) for a ∈ A, s0 (a3m+1 ) =
s0 (a3m+2 ) = B − 1, s0 (a3m+3 ) = B + 2, and B 0 = 3B. Then A0 , m0 , s0 , and B 0 define
an instance of Odd-3-Partition. It is easy to verify that this is a yes instance of Odd3-Partition if and only if I is a yes instance of 3-Partition. The transformation can be
Chapter 5. Sorting Strings by Global Transformations
141
performed in polynomial-time, therefore Odd-3-Partition must be NP-complete in the
strong sense. 2
Theorem 5.7.2 BD is NP-complete, even when t = 2.
Proof We prove this theorem using a transformation that is very similar to the one
used in the proof of Theorem 5.7.1.
BD is in NP because, given a sequence of block-interchanges, it can easily be checked
in polynomial-time that the sequence transforms S into T and has length at most
d. By Lemma 5.7.2 the proof is completed by demonstrating a pseudo-polynomial
transformation from Odd-3-Partition to BD. The transformation is now described.
Let I be an instance of Odd-3-Partition. From this instance construct an instance
3m−1 s(ai ) 3
0
B
8m−3 , and d =
I of BD with S = (++i=1
0
1 ) ++ 0s(a3m ) , T = (++m
i=1 0 1) ++ 1
(3m − 1)/2. So the blocks of zeros in S represent elements of A and the lengths of the
blocks represent the the sizes of the elements.
We now show that this transformation from Odd-3-Partition to BD is a pseudopolynomial transformation. There are four properties a), b), c) and d), that must be
verified, and because the transformation is so similar to the transformation used in
Theorem 5.7.1 properties b), c) and d) follow easily from the proof of that theorem.
For property a), we have to show that I is a yes instance of Odd-3-Partition if and
only if bd(S, T ) ≤ (3m − 1)/2.
First of all we prove that bd(S, T ) ≥ (3m − 1)/2.
As in the proof of Theorem 5.7.1, f (U ) = z(U ) − o(U ) − 1. We show that, for any
block-interchange that transforms U into U 0 , f (U 0 ) ≥ f (U ) − 2. This will prove that
bd(S, T ) ≥ (3m − 1)/2, since f (S) = 3m − 1, and f (T ) = 0.
Suppose ρ is a block-interchange that transforms U into U 0 such that f (U 0 ) < f (U ).
Then ρ must reduce the number of blocks of zeros, or increase the number of blocks of
ones of length one, or do both.
Suppose ρ is a transposition such that f (U 0 ) < f (U ). If ρ reduces the number of
blocks of zeros, then f (U 0 ) = f (U )−1 because a transposition can reduce the number of
blocks of zeros by at most one (Lemma 5.5.4), and such a transposition cannot increase
the number of blocks of ones of length one. If ρ increases the number of blocks of ones
of length one then f (U 0 ) ≥ f (U ) − 2, because a transposition can increase o by at most
two. If U does not contain any 0110 substrings then then f (U 0 ) = f (U ) − 1 because
then ρ can only increase o by one without increasing z, or it can increase o by two while
also increasing z by one, e.g., . . . 11[10 . . . 01][11 . . . 0]0 . . .. If U does contain a 0110
substring then f (U 0 ) ≥ f (U ) − 2, and this bound can be achieved by a transposition
that creates two blocks of ones of length one without changing the number of blocks
of zeros, e.g., . . . 01[10 . . . 111][0 . . . 0]11 . . ..
If ρ is not a transposition then ρ must have the form
. . . a[b . . . c]d . . . e[f . . . g]h . . . = . . . af . . . gd . . . eb . . . ch.
Chapter 5. Sorting Strings by Global Transformations
142
Note that changes at ab and ef happen independently of changes at cd and gh. Suppose that the number of blocks of zeros is decreased by the action of ρ at ab and
ef . Then ρ must have the form . . . 0[1 . . .] . . . 1[0 . . .] . . ., or alternatively the form
. . . 1[0 . . .] . . . 0[1 . . .] . . .. Therefore, f is decreased by one as a result of this part of the
block-interchange. Now suppose that the number of blocks of ones of length one is
increased by the action of ρ at ab and ef . Then ρ has the form:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
. . . 01[1 . . .] . . . 11[0 . . .] . . .,
. . . 11[0 . . .] . . . 01[1 . . .] . . .,
. . . 0[11 . . .] . . . 1[10 . . .] . . .,
. . . 1[10 . . .] . . . 0[11 . . .] . . .,
. . . 01[10 . . .] . . . 0[0 . . .] . . ., or
. . . 0[0 . . .] . . . 01[10 . . .] . . ..
In each case f is reduced by one as a result of this part of the block-interchange. Note
that other apparent possibilities, e.g., . . . 01[11 . . .] . . . 0[0 . . .] . . ., increase the number
of blocks of zeros and hence do not decrease f . The changes that can happen at cd
and gh are identical to those that can happen at ab and ef , therefore f can be reduced
overall by two by a single block-interchange. Hence bd(S, T ) ≥ (3m − 1)/2.
A block-interchange is bad if it splits a 00 substring or it is a transposition. Suppose
that bd(S, T ) = (3m − 1)/2. Then we show that no sequence of block-interchanges that
achieves this bound contains a bad block-interchanges. Clearly every block-interchange
in the sequence that transforms S into T must reduce f by two. For a block-interchange
in a minimal length sequence to be bad, the string must contain 0110 as a substring.
However S contains no such substring, and no block-interchange that reduces f by two
can create such a substring. So if bd(S, T ) = (3m − 1)/2, then no shortest sequence of
block-interchanges that transforms S into T contains a bad block-interchange.
Therefore, if bd(S, T ) = (3m − 1)/2, then any minimum length sequence of blockinterchanges must assemble each block of zeros in T out of three blocks of zeros in S.
So if bd(S, T ) = (3m − 1)/2 then I is a yes instance of Odd-3-Partition.
We now show that if I is a yes instance of Odd-3-Partition, then bd(S, T ) ≤ (3m −
1)/2. For each subset Ai use a block-interchange of the form . . . 0[1 . . . 1]0 . . . 1[0 ∼
0]1 . . . to merge the three blocks of zeros representing elements of A i into a single
block of zeros. Then use (m − 1)/2 block-interchanges of the form . . . 01[1 . . . 1]10 ∼
01 . . . 1[0 ∼ 0]1, to create the blocks of ones of length one separating the blocks of zeros.
This establishes the required property a). So, by Lemma 5.7.2, BD is NP-complete.
2
5.8
Conclusion
In this chapter we have shown that, just as sorting permutations by reversals is NPhard, so also is finding the reversal distance between two strings, even when the
143
Chapter 5. Sorting Strings by Global Transformations
strings are drawn from a binary alphabet. We have also shown that finding the blockinterchange distance between two strings is NP-hard. This contrasts with sorting permutations by block-interchanges, for which a polynomial-time algorithm was presented
in Chapter 4.
The complexity of finding the transposition distance between two strings remains
open, just as the complexity of sorting permutations by transpositions is open. The
complexity of finding the prefix-reversal distance between two strings (over a fixed
size alphabet) is also open, unlike the complexity of sorting permutations by prefixreversals, which is known to be NP-hard.
For all four problems on strings, we derived lower and upper bounds for the distance
between binary strings, and used these bounds to find the diameter of Bn .
A summary of the known complexity of the distance problems on strings (over a
fixed size alphabet) and the distance problems on permutations is presented in Table
5.2. The table also shows the diameter of Bn under each of the operations.
permutations
strings
diameter of Bn
Reversals
NP-hard
NP-hard
bn/2c
Prefix-Reversals
NP-hard
Open
n−1
Transpositions
Open
Open
bn/2c
Block-Interchanges
P
NP-hard
d(n − 1)/4e
Table 5.2: The complexity of the different problems, and the diameter of Bn .
Pevzner and Waterman discuss the problem of sorting strings by reversals (they
call the problem sorting words by reversals) in their open problems paper [PW95].
Their Problem 4 is to devise a performance guarantee algorithm for sorting words by
reversals. Finding an algorithm with a performance guarantee for any of the distance
problems on strings remains an open problem.
Glossary
Symbol
Bk
BRn
Bn
bb(S, T )
bD(n)
bD2 (n)
bd(S, T )
bd(π)
bi
b(S)
Ck
Cmax
Cmin
|C|
CM
C+
C · ρ(u)
fab (S)
gl(π)
g(v)
I(C)
i(C, k)
ı
ın
+ı
−ı
KM
L(C)
lb (u)
l(C)
lcp(S, T )
Short description
Bk = 0k ++ 1k
the broken permutation of length n
the set of binary strings of length n
the number of block-interchange breakpoints
the block-interchange diameter of Sn
the block-interchange diameter of Bn
the block-interchange distance between strings S and T
the block-interchange distance of π
the black edge (π(i), π(i − 1))
the number of blocks in S
Ck = (10)k
the position of the rightmost edge of cycle C
the position of the leftmost edge of cycle C
the number of cycles in C
a cycle decomposition derived from M
the augmented cycle decomposition derived from C
the cycle decomposition derived from C after applying ρ(u)
the number of occurrences of the substring ‘ab’ in S
the ‘glued’ permutation derived from π
the grey edge of rG0 (π) associated with v
the inner sequence of cycle C
the kth element of the inner sequence of C
the identity permutation
the identity permutation of length n
+ı = [+1 +2 . . . +n]
−ı = [−1 −2 . . . −n]
the subgraph of rR(CM ) that is related to kernel K
the canonical labelling of cycle C
the position of the leftmost black edge adjacent to g(u) in C
the length of cycle C
the length of the longest common prefix of S and T
144
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115
82
115
131
110
135
114
12
50
116
115
51
51
27
38
18
20
117
48
18
55
55
2
2
6
9
45
55
18
50
116
145
Glossary
lcs(S, T )
lg (u)
lis(π)
n
O(C)
o(C, k)
o(S)
prb(π)
prD(n)
prD2 (n)
prd(S, T )
prd(π)
p(π, u)
Rn
+Rn
R(π)
rb(S, T )
rb (u)
rb(π)
rc(π)
rc2 (π)
rc3∗ (π)
rD(n)
rD2 (n)
rd(S, T )
rd(π)
rF (π)
rF ∗ (π)
rf (π)
rG(π)
rG0 (π)
rg (u)
rh(π)
rL(M )
rR(C)
ru(C)
S
Sı
Sk
Sn
S −1
the length of the longest common suffix of S and T
the position of the leftmost element incident to g(u)
the length of the longest increasing subsequence of π
the length of a permutation
the outer sequence of cycle C
the kth element of the outer sequence of C
the number of blocks of ones of length one in S
the number of prefix reversal breakpoints in π
the prefix reversal diameter of Sn
the prefix reversal diameter of Bn
the prefix reversal distance between strings S and T
the prefix reversal distance of π
the position of the edge u in tG(π)
the reverse permutation of length n
+Rn = [+n +(n − 1) . . . +1]
set of 2-reversals on π
the number of reversal breakpoints
the position of the rightmost black edge incident to g(u) in C
the number of reversal breakpoints in π
the size of a maximum alternating cycle decomposition of rG(π)
the least number of 2-cycles in any maximum cycle decomposition
the number of cycles of length greater than 2
the reversal diameter of Sn
the reversal diameter of Bn
the reversal distance between strings S and T
the reversal distance of π
the matching graph
the bridge free subgraph of rF (π)
the fortress value of π
the reversal cycle graph of π
the augmented reversal cycle graph of π
the position of the rightmost element adjacent to g(u)
the number of hurdles in rG(π)
the ladder graph
the reversal graph based on C
the number of unoriented components in rR(C + )
the string S in reverse order
the string related to S that has the form 0 ∼ 01 ∼ 1
the string obtained by concatenating k copies of S
the symmetric group
the string obtained by swapping 1s and 0s in S
116
18
11
2
55
55
138
8
8
124
114
8
52
11
7
37
117
18
4
5
28
28
5
119
114
3
29
39
7
4
18
18
7
29
18
27
115
121
115
2
115
146
Glossary
sd(X, Y )
sprD(n)
sprd(π)
srD(n)
srd(π)
tb(S, T )
tb(π)
tc(π)
tcodd (π)
tceven (π)
tD(n)
tD2 (n)
td(S, T )
td(π)
tG(π)
tH(π)
th(π)
thd (π)
tld(A, B)
tl(π)
ui
z(S)
∼
++
(i1 , . . . , ik )
α
β(i, j, k, l)
∆f (π, τ )
κk
|π|
πC
π·γ
ρ(i, j)
ρ(u)
Σn
τ (i, j, k)
ω
ωr
the syntenic edit distance between X and Y
the signed prefix reversal diameter
the signed prefix reversal distance of π
the signed reversal diameter of Σn
the signed reversal distance of π
the number of transposition breakpoints
the number of transposition breakpoints in π
the number of alternating cycles in tG(π)
the number of odd length alternating cycles in tG(π)
the number of even length alternating cycles in tG(π)
the transposition diameter of Sn
the transposition diameter of Bn
the transposition distance between strings S and T
the transposition distance of π
the transposition cycle graph of π
the transposition cycle overlap graph of π
the number of hurdles in tG(π)
the number of detectable hurdles in tG(π)
the reciprocal translocation distance between A and B
a lower bound for td(π)
a vertex of rR(C)
the number of blocks of zeros in S
a symbol used to represent repetition in strings
concatenation operator
an alternating cycle
a special character used to represent the start of a string
a block-interchange
The change in f as a result of applying τ to π
a permutation with k knots in tG(π)
the length of π
the component permutation
the permutation obtained by applying γ to π
a reversal
the reversal represented by u
the set containing all signed permutations of length n
a transposition
a special character used to represent the end of a string
the permutation of length 2r whose cycle graph is a knot
14
9
8
7
6
127
10
50
10
50
10
129
114
10
10
77
78
79
14
79
18
116
116
34
51
117
107
48
79
2
78
3
3
20
7
9
117
58
Bibliography
[ABIR89]
N. Amato, M. Blum, S. Irani, and R. Rubinfeld. Reversing trains: A turn
of the century sorting problem. Journal of Algorithms, 10:413–428, 1989.
[AD86]
D. Aldous and P. Diaconis. Shuffling cards and stopping times. American
Mathematical Monthly, 93(5):333–348, 1986.
[AL90]
A. Aggarwal and T. Leighton. A tight lower bound for the train reversal
problem. Information Processing Letters, 35:301–304, 1990.
[AW87]
M. Aigner and D. B. West. Sorting by insertion of leading elements. Journal of Combinatorial Theory, Series A, 45:306–309, 1987.
[BH96]
P. Berman and S. Hannenhalli. Fast sorting by reversals. In Combinatorial
Pattern Matching, Proceedings of the 7th Annual Symposium (CPM’96),
number 1075 in Lecture Notes in Computer Science, pages 168 – 185, 1996.
[BKS96]
M. Blanchette, T. Kunisawa, and D. Sankoff. Parametric genome rearrangement. Gene, 172:GC 11–GC 17, 1996.
[BP93]
V. Bafna and P. A. Pevzner. Genome rearrangements and sorting by
reversals. In Proceedings of the 34th IEEE Symposium on Foundations of
Computer Science, pages 148–157, 1993.
[BP95a]
V. Bafna and P. A. Pevzner. Sorting by reversals: Genome rearrangements
in plant organelles and evolutionary history of X chromosome. Molecular
Biology and Evolution, 12(2):239–246, 1995.
[BP95b]
V. Bafna and P. A. Pevzner. Sorting by transpositions. In Proceedings
of the 6th Annual ACM-SIAM Symposium on Discrete Algorithms, pages
614–623, 1995.
[BP96]
V. Bafna and P. A. Pevzner. Genome rearrangements and sorting by
reversals. SIAM Journal on Computing, 25(2):272–289, 1996.
[BP98]
V. Bafna and P. A. Pevzner. Sorting by transpositions. SIAM Journal on
Discrete Mathematics, 11(2):224–240, 1998.
147
Bibliography
148
[Cap97a]
A. Caprara. Formulations and complexity of multiple sorting by reversals.
Technical Report OR-97-15, DEIS - Operations Research Group, University of Bologna, 1997.
[Cap97b]
A. Caprara. Sorting by reversals is difficult. In Proceedings of the
First International Conference on Computational Molecular Biology (RECOMB’97), pages 75–83. ACM Press, 1997.
[Cap97c]
A. Caprara. Sorting permutations by reversals and eulerian cycle decompositions. Technical Report OR-97-4, DEIS - Operations Research Group,
University of Bologna, 1997.
[Cap98]
A. Caprara. On the tightness of the alternating-cycle lower bound for sorting by reversals. Technical Report OR-98-4, DEIS - Operations Research
Group, University of Bologna, 1998.
[CB95]
D. S. Cohen and M. Blum. On the problem of sorting burnt pancakes.
Discrete Applied Mathematics, 61:105–120, 1995.
[Chr96]
D. A. Christie. Sorting permutations by block-interchanges. Information
Processing Letters, 60(4):165–169, 1996.
[Chr98]
D. A. Christie. A 3/2-approximation algorithm for sorting by reversals. In
Proceedings of the 9th Annual ACM-SIAM Symposium on Discrete Algorithms, pages 244–252, 1998.
[CKb]
Just be.
[CLN95]
A. Caprara, G. Lancia, and S. K. Ng. A column-generation based branchand-bound algorithm for sorting by reversals. Presented at the DIMACS
4th Annual Implementation Challenge Workshop, 1995. Also Technical Report OR-95-10, DEIS - Operations Research Group, University of Bologna.
[CS95]
T. Chen and S. S. Skiena. Sorting with fixed-length reversals. Technical
Report 95-18, DIMACS, June 1995.
[CS96]
T. Chen and S. S. Skiena. Sorting with fixed-length reversals. Discrete
Applied Mathematics, 71:269–295, 1996.
[Dew87]
A. K. Dewdney. Computer recreations. Scientific American, 256(6):128–
129, 1987.
[DJK+ 97]
B. DasGupta, T. Jiang, S. Kannan, M. Li, and Z. Sweedyk. On the
complexity and approximation of syntenic distance. In Proceedings of the
1st Annual International Conference on Computational Molecular Biology
(RECOMB’97), pages 99–108, 1997.
Bibliography
149
[DMP95]
P. Diaconis, M. McGrath, and J. Pitman. Riffle shuffles, cycles, and descents. Combinatorica, 15:11–29, 1995.
[Dwe75]
H. Dweighter. American Mathematical Monthly, 82:1010, 1975.
[EG81]
S. Even and O. Goldreich. The minimum-length generator sequence problem is NP-hard. Journal of Algorithms, 2:311–313, 1981.
[Eid86]
J. A. Eidswick. Cubelike puzzles—what are they and how do you solve
them? American Mathematical Monthly, 93(3):157–176, 1986.
[FHL80]
M. Furst, J. Hopcroft, and E. Luks. Polynomial-time algorithms for permutation groups. In Proceedings of the 21st IEEE Symposium on Foundations
of Computer Science, pages 36–41, 1980.
[FNS96]
V. Ferretti, J. H. Nadeau, and D. Sankoff. Original synteny. In Combinatorial Pattern Matching, Proceedings of the 7th Annual Symposium
(CPM’96), number 1075 in Lecture Notes in Computer Science, pages 159
– 167, 1996.
[Fri98]
T. Frings. Approximationsalgorithmen für das Problem “Sorting by Reversals” auf Permutationen. Diplomarbeit, Rheinische Friedrich-WilhelmsUniversität Bonn, March 1998.
[GHV95]
S. A. Guyer, L. S. Heath, and J. P. C. Vergara. Subsequence and run
heuristics for sorting by transpositions. Presented at the DIMACS 4th
Annual Algorithm Implementation Challenge, 1995. Also Technical Report
TR-97-20, Virginia Tech Department of Computer Science.
[GJ75]
M. R. Garey and D. S. Johnson. Complexity results for multiprocessor
scheduling under resource constraints. SIAM Journal of Computing, 4:397–
411, 1975.
[GJ79]
M. R. Garey and D. S. Johnson. Computers and Intractability: A Guide
to the Theory of NP-Completeness. W. H. Freeman and Company, New
York, 1979.
[GP79]
W. H. Gates and C. H. Papadimitriou. Bounds for sorting by prefix reversal. Discrete Mathematics, 27:47–57, 1979.
[GPIC97]
Q.-P. Gu, S. Peng, K. Iwata, and Q. M. Chen. A heuristic algorithm
for genome rearrangements. In Proceedings of the Genome Informatics
Workshop, Tokyo, 1997. Universal Academy Press. http://www.tokyocenter.genome.ad.jp/manuscripts/GIW97/Poster/GIW97P08.pdf.
Bibliography
150
[GPS96]
Q.-P. Gu, S. Peng, and H. Sudborough. Approximation algorithms
for genome rearrangements. In Proceedings of the Genome Informatics
Workshop, Tokyo, 1996. Universal Academy Press. http://www.tokyocenter.genome.ad.jp/manuscripts/GIW96/Oral/GIW96O02.ps.
[GPS99]
Q.-P. Gu, S. Peng, and H. Sudborough. A 2-approximation algorithm
for genome rearrangements by reversals and transpositions. Theoretical
Computer Science, 210(2):327–339, 1999.
[Gus97]
D. Gusfield. Algorithms On Strings, Trees and Sequences: Computer Science and Computational Biology. Cambridge University Press, Cambridge,
1997.
[Han95a]
S. Hannenhalli. Polynomial-time algorithm for computing translocation
distance between genomes. In Combinatorial Pattern Matching, Proceedings of the 6th Annual Symposium (CPM’95), number 937 in Lecture Notes
in Computer Science, pages 162–176, 1995.
[Han95b]
S. Hannenhalli. Transforming men into mice (a computational theory of
genome rearrangements). PhD thesis, Pennsylvania State University Department of Computer Science and Engineering, 1995.
[Han96]
S. Hannenhalli. Polynomial-time algorithm for computing translocation
distance between genomes. Discrete Applied Mathematics, 71:137–151,
1996.
[HCKP95] S. Hannenhalli, C. Chappey, E. V. Koonin, and P. A. Pevzner. Genome
sequence comparison and scenarios for gene rearrangements: A test case.
Genomics, 30:299–311, 1995.
[HP95a]
S. Hannenhalli and P. A. Pevzner. Towards a computational theory of
genome rearrangements. Lecture Notes in Computer Science, 1000:184–
202, 1995.
[HP95b]
S. Hannenhalli and P. A. Pevzner. Transforming cabbage into turnip (polynomial algorithm for sorting signed permutations by reversals). In Proceedings of the 27th Annual ACM Symposium on Theory of Computing,
pages 178–189, 1995.
[HP95c]
S. Hannenhalli and P. A. Pevzner. Transforming men into mice (polynomial algorithm for genomic distance problem). In Proceedings of
the 36th Annual IEEE Symposium on Foundations of Computer Science
(FOCS’95), pages 581–592, 1995.
[HP96]
S. Hannenhalli and P. A. Pevzner. To cut . . . or not to cut (applications of
comparative physical maps in molecular evolution). In Proceedings of the
Bibliography
151
7th Annual ACM-SIAM Symposium on Discrete Algorithms, pages 304–
313, 1996.
[HS]
M. H. Heydari and I. H. Sudborough. Pancake sorting is NP-complete.
manuscript.
[HS97]
M. H. Heydari and I. H. Sudborough. On the diameter of the pancake
network. Journal of Algorithms, 25:67–94, 1997.
[HV95]
L. S. Heath and J. P. C. Vergara. Some experiments on the sorting by
reversals problem. Technical Report TR-95-16, Virginia Tech Department
of Computer Science, September 1995.
[HV97]
L. S. Heath and J. P. C. Vergara. Sorting by bounded block-moves. Technical Report TR-97-09, Virginia Tech Department of Computer Science,
May 1997. To appear in Discrete Applied Mathematics.
[HV98]
L. S. Heath and J. P. C. Vergara. Sorting by short block-moves. Technical Report TR-98-03, Virginia Tech Department of Computer Science,
February 1998.
[IC95]
R. W. Irving and D. A. Christie. Sorting by reversals: on a conjecture
of Kececioglu and Sankoff. Technical Report TR-95-12, Department of
Computing Science of The University of Glasgow, May 1995.
[Jer85]
M. R. Jerrum. The complexity of finding minimum-length generator sequences. Theoretical Computer Science, 36:265–289, 1985.
[Jor95]
E. M. Jordan. Visualising measures of genetic distance. Presented at
the DIMACS 4th Annual Implementation Challenge Workshop, 1995.
http://medg.lcs.mit.edu/people/emjordan/final.ps.
[Knu73]
D. E. Knuth. Sorting and Searching, volume 3 of The Art of Computer Programming. Addison-Wesley, Reading, Massachusetts, first edition, 1973.
[Knu98]
D. E. Knuth. Sorting and Searching, volume 3 of The Art of Computer
Programming. Addison-Wesley, Reading, Massachusetts, second edition,
1998.
[KR95]
J. D. Kececioglu and R. Ravi. Of mice and men: Algorithms for evolutionary distances between genomes with translocation. In Proceedings
of the 6th Annual ACM-SIAM Symposium on Discrete Algorithms, pages
604–613, 1995.
[KS93]
J. Kececioglu and D. Sankoff. Exact and approximation algorithms for the
inversion distance between two chromosomes. In Combinatorial Pattern
Bibliography
152
Matching, Proceedings of the 4th Annual Symposium (CPM’93), volume
684 of Lecture Notes in Computer Science, pages 87–105, 1993.
[KS94]
J. D. Kececioglu and D. Sankoff. Efficient bounds for oriented chromosome
inversion distance. In Combinatorial Pattern Matching, Proceedings of
the 5th Annual Symposium (CPM’94), volume 807 of Lecture Notes in
Computer Science, pages 307–325, 1994.
[KS95]
J. Kececioglu and D. Sankoff. Exact and approximation algorithms for
sorting by reversals, with application to genome rearrangement. Algorithmica, 13:180–210, 1995.
[KST97]
H. Kaplan, R. Shamir, and R. E. Tarjan. Faster and simpler algorithm for
sorting signed permutations by reversals. In Proceedings of the 8th Annual
ACM-SIAM Symposium on Discrete Algorithms, pages 344–351, 1997.
[LP86]
L. Lovász and M. D. Plummer. Annals of Discrete Mathematics (29):
Matching Theory. Number 121 in North-Holland Mathematics Studies.
North-Holland, Amsterdam, 1986.
[LW75]
R. Lowrance and R. A. Wagner. An extension of the string-to-string correction problem. Journal of the Association for Computing Machinery,
22(2):177–183, April 1975.
[MWD97a] J. Meidanis, M. E. M. T. Walter, and Z. Dias. Distância de reversão de
cromossomos cirulares com sinais. In Proceedings of the 14th SEMISH Software and Hardware Seminar, 1997.
[MWD97b] J. Meidanis, M. E. M. T. Walter, and Z. Dias. Transposition distance
between a permutation and its reverse. In Proceedings of the 4th South
American Workshop on String Processing, pages 70–79, 1997.
[Pap94]
C. Papadimitriou. Computational Complexity. Addison-Wesley, Reading,
Massachusetts, 1994.
[PW95]
P. A. Pevzner and M. S. Waterman. Open combinatorial problems in computational molecular biology. In Proceedings of the 3rd Israel Symposium
on the Theory of Computing and Systems, pages 158–173, 1995.
[SM97]
J. Setubal and J. Meidanis. Introduction to Computational Molecular Biology. PWS Publishing Company, Boston, 1997.
[Tra97]
N. Tran. An easy case of sorting by reversals. In Combinatorial Pattern Matching, 8th Annual Symposium, volume 1264 of Lecture Notes in
Computer Science, pages 83–89. Springer, 1997.
Bibliography
153
[Ver97]
J. P. C. Vergara. Sorting by Bounded Permutations. PhD thesis, Virginia
Tech Department of Computer Science, April 1997.
[Wag83]
R. A. Wagner. On the complexity of the extended string-to-string correction problem. In D. Sankoff and J. B. Kruskal, editors, Time Warps, String
Edits and Macromolecules: The Theory and Practice of Sequence Comparison, pages 215–235. Addison-Wesley, Reading, Massachusetts, 1983.
[WDM98]
M. E. M. T. Walter, Z. Dias, and J. Meidanis. Reversal and transposition
distance of linear chromosomes. To be presented at SPIRE’98 - String
Processing and Information Retrieval: A South American Symposium,
September 1998.
[WEHM82] G. A. Watterson, W. J. Ewens, T. E. Hall, and A. Morgan. The chromosome inversion problem. Journal of Theoretical Biology, 99:1–7, 1982.
[WF74]
R. A. Wagner and M. J. Fischer. The string-to-string correction problem. Journal of the Association for Computing Machinery, 21(1):168–173,
January 1974.